ORCCA Substitution

Section 4.2 Substitution

Objectives: PCC Course Content and Outcome Guide

In Section 1, we used graphing to solve a system of two linear equations. While graphing is insightful, it can be time-consuming. And what happens if the actual solution to the system does not have whole number coordinates—will you be able to see exactly what are the coordinates of the crossing? And what if your hand-drawn graph has some small imperfection that makes it look like the lines are crossing at the wrong place?

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There are two main alternatives to graphing that can find the solution to a system of two linear equations. They use pencil-and-paper algebra instead of a picture. This section covers one of those alternative methods, and the next section covers the other.

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Figure 4.2.1. Alternative Video Lesson

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Subsection 4.2.1 Substitution

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Example 4.2.2. The Interview.

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Once upon a time, the New York Times

nyti.ms/2pupebT

published an article about the movie, The Interview. It included the following quote:

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The Interview generated roughly $$ 15$ million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.

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Sony did not say how much of that total represented $$ 6$ digital rentals versus $$ 15$ sales. The studio said there were about two million transactions overall.

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Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog

www.joeydevilla.com/2014/12/31/

that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)

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First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let

r

be the number of rental transactions and let

s

be the number of sales transactions.

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If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both

r

and

s

are measured in “transactions”. The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is:

(r transactions) + (s transactions) = 2,000,000 transactions

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Without the units:

r + s = 2,000,000

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The price of each rental was

$ 6 .

That means the problem has given us a rate of

6 \frac{dollars}{transaction}

to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by

r,

and that would give us the total number of dollars generated from rentals. This is

6 r .

Similarly, the price of each sale was

$ 15,

so the revenue from sales was

15 s .

The total revenue was

$ 15

million, which we can represent with this equation:

(6 \frac{dollars}{transaction}) (r transactions) + (15 \frac{dollars}{transaction}) (s transactions) = $ 15,000,000

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Without the units:

6 r + 15 s = 15,000,000

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Here is our system of equations, with the commas removed:

{\begin{aligned} r & + & s & = & 2 000 000 \\ 6 r & + & 15 s & = & 15 000 000 \end{aligned}

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To solve the system, we will use a method called substitution. The idea is to use one equation to isolate

r .

Then substitute this for the “

r

” that’s in the other equation. This leaves you with one equation where the only variable is

s .

And we can handle that directly.

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The first equation from the system is an easy one to isolate

r :

\begin{aligned} r + s & = 2 000 000 & (the system's first equation) \\ r & = 2 000 000 - s \end{aligned}

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This tells us that the expression

2 000 000 - s

is equal to

r,

so we can substitute that in place of

r

in the second equation:

\begin{aligned} 6 r + 15 s & = 15 000 000 & (the system's second equation) \\ 6 (2 000 000 - s) + 15 s & = 15 000 000 \end{aligned}

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Now we have an equation with only one variable,

s .

\begin{aligned} 6 (2 000 000 - s) + 15 s & = 15 000 000 \\ 12 000 000 - 6 s + 15 s & = 15 000 000 \\ 12 000 000 + 9 s & = 15 000 000 \\ 9 s & = 3 000 000 \\ \frac{9 s}{9} & = \frac{3 000 000}{9} \\ s & = 333 333. \overset{―}{3} \end{aligned}

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At this point, we know that

s = 333 333. \overset{―}{3} .

This tells us that out of the

2

million transactions, roughly

333,333

were from online sales. Recall that we isolated

r

previously and found

r = 2 000 000 - s .

\begin{aligned} r & = 2 000 000 - s \\ r & = 2 000 000 - 333 333. \overset{―}{3} \\ r & = 1 666 666. \overset{―}{6} \end{aligned}

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In summary, there were roughly

333,333

sales and roughly

1,666,667

rentals.

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Remark 4.2.3.

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In Example 2, we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like

s = 330,000

and

r = 1,670,000

as long as we communicate clearly that we rounded.

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In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values.

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Example 4.2.4.

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Solve the system of equations using substitution:

\begin{array}{r} {\begin{aligned} x & + & 2 y & = & 5 \\ 3 x & - & 2 y & = & 8 \end{aligned} \end{array}

Explanation.

To use substitution, we need to isolate one of the variables in one of our equations. Looking over both equations, it will be easiest to isolate

x

in the first equation:

\begin{aligned} x + 2 y & = 5 \\ x & = 5 - 2 y \end{aligned}

Next, we substitute

5 - 2 y

in for

x

in the second equation, giving us a linear equation in only one variable

y .

And this is an equation that we may solve using skills from Section 2.1.

\begin{aligned} 3 x - 2 y & = 8 \\ 3 (5 - 2 y) - 2 y & = 8 \\ 15 - 6 y - 2 y & = 8 \\ 15 - 8 y & = 8 \\ - 8 y & = - 7 \\ y & = \frac{7}{8} \end{aligned}

Now that we have the value for

y,

we need to find the value for

x .

We already isolated

x,

and it’s easiest to just use that equation.

\begin{aligned} x & = 5 - 2 y \\ x & = 5 - 2 (\frac{7}{8}) \\ x & = 5 - \frac{7}{4} \\ x & = \frac{20}{4} - \frac{7}{4} = \frac{13}{4} \end{aligned}

At this point we think the solution is the point

(\frac{13}{4}, \frac{7}{8})

or in other words:

x = \frac{13}{4},

y = \frac{7}{8} .

It’s only human to make mistakes though, so we should check that the solution actually works. To check it, try using

x = \frac{13}{4},

y = \frac{7}{8}

in both of the original equations.

\begin{aligned} x + 2 y & = 5 & 3 x - 2 y & = 8 \\ \frac{13}{4} + 2 (\frac{7}{8}) & \overset{?}{=} 5 & 3 (\frac{13}{4}) - 2 (\frac{7}{8}) & \overset{?}{=} 8 \\ \frac{13}{4} + \frac{7}{4} & \overset{?}{=} 5 & \frac{39}{4} - \frac{7}{4} & \overset{?}{=} 8 \\ \frac{20}{4} & \overset{✓}{=} 5 & \frac{32}{4} & \overset{✓}{=} 8 \end{aligned}

We conclude then that this system of equations is true when

x = \frac{13}{4}

and

y = \frac{7}{8} .

The solution is the point

(\frac{13}{4}, \frac{7}{8})

and we write the solution set as

{(\frac{13}{4}, \frac{7}{8})} .

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Checkpoint 4.2.5.

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Use substitution to solve the system.

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\begin{array}{r} {\begin{aligned} x & - & 3 y & = & 1 \\ 2 x & + & y & = & - 3 \end{aligned} \end{array}

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(a)

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Isolate one of the variables in one of the equations.

Explanation.

We choose to isolate

x

in the first equation. All we need to do is add

3 y

to each side, and then

x = 3 y + 1 .

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(b)

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Substitute the isolated expression into the other equation.

Explanation.

Making the substitution:

2 (3 y + 1) + y = - 3 .

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(c)

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Solve for the one variable that remains.

Explanation.

\begin{aligned} 2 (3 y + 1) + y & = - 3 \\ 6 y + 2 + y & = - 3 \\ 7 y + 2 & = - 3 \\ 7 y & = - 5 \\ y & = - \frac{5}{7} \end{aligned}

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(d)

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Solve for the other variable.

Explanation.

\begin{aligned} x & = 3 y + 1 \\ x & = 3 (- \frac{5}{7}) + 1 \\ x & = - \frac{15}{7} + \frac{7}{7} \\ x & = - \frac{8}{7} \end{aligned}

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Sometimes it makes more sense to start the process using the second equation instead of the first. And sometimes it makes more sense to isolate the second variable instead of the first.

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Example 4.2.6.

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Solve this system of equations using substitution:

\begin{array}{r} {\begin{aligned} 3 x & - & 7 y & = & 5 \\ - 5 x & + & 2 y & = & 11 \end{aligned} \end{array}

Explanation.

We need to isolate one of the variables in one of the equations. Looking over both equations, it will be easiest to isolate

y

in the second equation because that is where the coefficient is smallest. We are going to have to divide by some coefficient; it might as well be the smallest one to keep the fraction arithmetic as simple as we can.

\begin{aligned} - 5 x + 2 y & = 11 \\ 2 y & = 11 + 5 x \\ \frac{2 y}{2} & = \frac{11 + 5 x}{2} \\ y & = \frac{11}{2} + \frac{5}{2} x \end{aligned}

Note that there are fractions once we’ve isolated

y .

We should take care with the steps that follow to make sure that the fraction arithmetic is correct.

Substitute

\frac{11}{2} + \frac{5}{2} x

in for

y

in the first equation, and that leads to having a single linear equation with only one variable. We can solve this as we did in Section 2.3. Note the step in the middle where we clear denominators.

\begin{aligned} 3 x - 7 y & = 5 \\ 3 x - 7 (\frac{11}{2} + \frac{5}{2} x) & = 5 \\ 3 x - \frac{77}{2} - \frac{35}{2} x & = 5 \\ 2 \cdot (3 x - \frac{77}{2} - \frac{35}{2} x) & = 2 \cdot (5) \\ 6 x - 77 - 35 x & = 10 \\ - 29 x - 77 & = 10 \\ - 29 x & = 87 \\ x & = \frac{87}{- 29} = - 3 \end{aligned}

Now that we have the value for

x,

we need to find the value for

y .

We already isolated

y,

and it’s easiest to just use that equation.

\begin{aligned} y & = \frac{11}{2} + \frac{5}{2} x \\ y & = \frac{11}{2} + \frac{5}{2} (- 3) \\ y & = \frac{11}{2} - \frac{15}{2} \\ y & = - \frac{4}{2} = - 2 \end{aligned}

To check the solution we think we’ve found, try using

x = - 3,

y = - 2

in both of the original equations.

\begin{aligned} 3 x - 7 y & = 5 & - 5 x + 2 y & = 11 \\ 3 (- 3) - 7 (- 2) & \overset{?}{=} 5 & - 5 (- 3) + 2 (- 2) & \overset{?}{=} 11 \\ - 9 + 14 & \overset{✓}{=} 5 & 15 - 4 & \overset{✓}{=} 11 \end{aligned}

We conclude then that this system of equations is true when

x = - 3

and

y = - 2 .

The solution is the point

(- 3, - 2)

and we write the solution set as

{(- 3, - 2)} .

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A system may start out with fractions among the coefficients. Just as we learned in Section 2.3, the algebra can go more smoothly if we clear the denominators before doing more work.

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Example 4.2.7.

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Solve the system of equations using the substitution method.

\begin{array}{r} {\begin{aligned} \frac{x}{3} - \frac{1}{2} y & = \frac{5}{6} \\ \frac{1}{4} x & = \frac{y}{2} + 1 \end{aligned} \end{array}

Explanation.

When a system of equations has fraction coefficients, it may be helpful to “clear the denominators”. With each equation, multiply each side by the least common multiple of that equation’s denominators.

In the first equation, the least common multiple of the denominators is

6,

so:

\begin{aligned} \frac{x}{3} - \frac{1}{2} y & = \frac{5}{6} \\ 6 \cdot (\frac{x}{3} - \frac{1}{2} y) & = 6 \cdot \frac{5}{6} \\ 2 x - 3 y & = 5 \end{aligned}

In the second equation, the least common multiple of the denominators is

4,

so:

\begin{aligned} \frac{1}{4} x & = \frac{y}{2} + 1 \\ 4 \cdot \frac{1}{4} x & = 4 \cdot (\frac{y}{2} + 1) \\ x & = 2 y + 4 \end{aligned}

Now we have a system that is equivalent to the original system of equations, but there are no fraction coefficients:

\begin{array}{r} {\begin{aligned} 2 x - 3 y & = 5 \\ x & = 2 y + 4 \end{aligned} \end{array}

The second equation has already isolated

x,

so we will substitute

2 y + 4

in for

x

in the first equation.

\begin{aligned} 2 x - 3 y & = 5 \\ 2 (2 y + 4) - 3 y & = 5 \\ 4 y + 8 - 3 y & = 5 \\ y + 8 & = 5 \\ y & = - 3 \end{aligned}

And we have solved for

y .

To find

x,

we know

x = 2 y + 4,

so we have:

\begin{aligned} x & = 2 y + 4 \\ x & = 2 (- 3) + 4 \\ x & = - 6 + 4 = - 2 \end{aligned}

The solution is

(- 2, - 3) .

(This should be checked in the original two equations though.)

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Real world applications can lead to a system where both equations are either in point-slope form or slope-intercept form. This is helpful, since it means one variable has already been isolated.

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Example 4.2.8.

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Solve the system of equations using the substitution method.

\begin{array}{r} {\begin{aligned} y & = \frac{5}{6} x + 3 \\ y & = \frac{2}{3} x + 1 \end{aligned} \end{array}

Explanation.

In the first equation,

y

is already isolated. So we can substitute

\frac{5}{6} x + 3

in for

y

in the second equation:

\frac{5}{6} x + 3 = \frac{2}{3} x + 1 .

Some people prefer to think of this as setting the two right sides equal to each other. That works too: if

\frac{5}{6} x + 3

is equal to

y,

and

\frac{2}{3} x + 1

is also equal to

y,

then these expressions are equal to each other. Either way:

\begin{aligned} \frac{5}{6} x + 3 & = \frac{2}{3} x + 1 \\ 6 \cdot (\frac{5}{6} x + 3) & = 6 \cdot (\frac{2}{3} x + 1) \\ 5 x + 18 & = 4 x + 6 \\ x + 18 & = 6 \\ x & = - 12 \end{aligned}

And then:

\begin{aligned} y & = \frac{5}{6} (- 12) + 3 \\ y & = - 10 + 3 = - 7 \end{aligned}

And the solution is

(- 12, - 7) .

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For summary reference, here is the general procedure.

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Process 4.2.9. Solving Systems of Equations by Substitution.

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To solve a system of two linear equations in two variables by substitution,

If helpful, clear denominators from each equation using the least common multiple of each equation’s denominators.
Isolate one variable in one of the equations.
Substitute that expression into the other equation. There should now only be one variable in that equation.
Solve that equation for the one remaining variable.
Substitute that value into the earlier result where the first variable was isolated, and solve for that variable.
Verify your solution using the original two equations.

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Subsection 4.2.2 Applications

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Example 4.2.10. Two Different Interest Rates.

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Notah made some large purchases with his two credit cards one month and took on a total of

$ 8400

in debt from the two cards. He didn’t make any payments the first month, so the two credit card debts each started accruing interest. That month, his Visa card charged

2 %

interest and his Mastercard charged

2.5 %

interest. Because of this, Notah’s total debt grew by

$ 178 .

How much money did Notah charge to each card?

Explanation.

We start by clearly defining two variables for the two unknowns. Let

V

be the amount charged to the Visa card (in dollars) and let

M

be the amount charged to the Mastercard (in dollars).

To set up the equations, notice that we are given two different total dollar amounts. One is the total debt Notah initially took on,

$ 8400 .

So we have:

(V dollars) + (M dollars) = $ 8400

Or without units:

V + M = 8400

The other total we were given is the total amount of interest,

$ 178 .

The Visa had

V

dollars charged to it and accrues

2 %

interest. So

0.02 V

is the dollar amount of interest that comes from using the Visa card. Similarly,

0.025 M

is the dollar amount of interest from using the Mastercard. Together:

0.02 (V dollars) + 0.025 (M dollars) = $ 178

Or without units:

0.02 V + 0.025 M = 178

As a system, we write:

{\begin{aligned} V & + & M & = 8400 \\ 0.02 V & + & 0.025 M & = 178 \end{aligned}

To solve this system by substitution, notice that it will be easier to isolate one of the variables in the first equation. We’ll isolate

V :

\begin{aligned} V + M & = 8400 \\ V & = 8400 - M \end{aligned}

Now we substitute

8400 - M

in for

V

in the second equation:

\begin{aligned} 0.02 V + 0.025 M & = 178 \\ 0.02 (8400 - M) + 0.025 M & = 178 \\ 168 - 0.02 M + 0.025 M & = 178 \\ 168 + 0.005 M & = 178 \\ 0.005 M & = 10 \\ \frac{0.005 M}{0.005} & = \frac{10}{0.005} \\ M & = 2000 \end{aligned}

And then we can determine the value of

V

by using the earlier equation where we isolated

V :

\begin{aligned} V & = 8400 - M \\ V & = 8400 - 2000 \\ V & = 6400 \end{aligned}

In summary, Notah charged

$ 6400

to the Visa and

$ 2000

to the Mastercard. We should check that these numbers work as solutions to our original system and that they make sense in context. (For instance, if one of these numbers were negative, or was something small like

$ 0.50,

they wouldn’t make sense as credit card debt.)

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The next two examples are called mixture problems, because they involve mixing two quantities together to form a combination and we want to find out how much of each quantity to mix.

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Example 4.2.11. Mixing Solutions with Two Different Concentrations.

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LaVonda is a meticulous bartender and she needs to serve

600

milliliters of Rob Roy, a cocktail that is

34 %

alcohol by volume. The main ingredients are scotch that is

42 %

alcohol and vermouth that is

18 %

alcohol. How many milliliters of each ingredient should she mix together to make the concentration she needs?

Explanation.

The two unknowns are the quantities of each ingredient. Let

S

be the amount of scotch (in mL) and let

V

be the amount of vermouth (in mL).

One quantity given to us in the problem is 600 mL. Since this is the total volume of the mixed drink, we must have:

(S mL) + (V mL) = 600 mL

Or without units:

S + V = 600

To build the second equation, we have to think about the alcohol concentrations for the pure scotch, pure vermouth, and the mixed Rob Roy. It can be tricky to think about percentages like these correctly. One strategy is to focus on the amount of alcohol (in mL). If we have

S

milliliters of scotch that is

42 %

alcohol, then

0.42 S

is the actual amount of alcohol (in mL) in that scotch. Similarly,

0.18 V

is the amount of alcohol in the vermouth. And the final cocktail is 600 mL of liquid that is

34 %

alcohol. So it has

0.34 (600) = 204

milliliters of alcohol. This is telling us:

0.42 (S mL) + 0.18 (V mL) = 204 mL

Or without units:

0.42 S + 0.18 V = 204

So our system is:

{\begin{aligned} S & + & V & = 600 \\ 0.42 S & + & 0.18 V & = 204 \end{aligned}

To solve this system, we’ll isolate

S

in the first equation:

\begin{aligned} S + V & = 600 \\ S & = 600 - V \end{aligned}

And then substitute

600 - V

in for

S

in the second equation with :

\begin{aligned} 0.42 S + 0.18 V & = 204 \\ 0.42 (600 - V) + 0.18 V & = 204 \\ 252 - 0.42 V + 0.18 V & = 204 \\ 252 - 0.24 V & = 204 \\ - 0.24 V & = - 48 \\ \frac{- 0.24 V}{- 0.24} & = \frac{- 48}{- 0.24} \\ V & = 200 \end{aligned}

Now we determine

S

using the equation where we had isolated

S :

\begin{aligned} S & = 600 - V \\ S & = 600 - 200 \\ S & = 400 \end{aligned}

In summary, LaVonda needs to combine 400 mL of scotch with 200 mL of vermouth to create 600 mL of Rob Roy.

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Let’s take a moment to consider estimation to ask whether the solution to Example 11 is reasonable. LaVonda will mix scotch (

42 %

concentration) with vermouth (

18 %

concentration) and wants to end with a

34 %

concentration. Is

34

closer to

42

18 ?

It’s closer to

42

so we should expect there to be more scotch than vermouth. This agrees with the solution we found.

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Checkpoint 4.2.12. Mixing a Coffee Blend.

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Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for

$ 12.50

per pound. They have some coffee beans from Columbia that sell for

$ 9.00

per pound and some coffee beans from Honduras that sell for

$ 14.00

per pound. How many pounds of each should they mix to make

30

pounds of the blend?

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(a)

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Write two equations that form a system for this scenario.

Explanation.

We have

C

pounds of Columbian coffee and

H

pounds of Honduran coffee. Since there must be

30

pounds total, one equation is

C + H = 30 .

The total cost of the Colombian coffee in the blend will be

9 C,

and the total cost of the Honduran coffee in the blend will be

14 H .

All together, the blend has a total cost of

12.50 \cdot 30 = 375

dollars. So another equation is

9 C + 14 H = 375 .

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(b)

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Isolate one of the variables in one of the equations.

Explanation.

We choose to isolate

C

in the first equation. All we need to do subtract

H

from each side, and then

C = 30 - H .

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(c)

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Substitute the isolated expression into the other equation.

Explanation.

Making the substitution:

9 (30 - H) + 14 H = 375 .

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(d)

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Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use.

Explanation.

\begin{aligned} 9 (30 - H) + 14 H & = 375 \\ 270 - 9 H + 14 H & = 375 \\ 270 + 5 H & = 375 \\ 5 H & = 105 \\ H & = 21 \end{aligned}

\begin{aligned} C & = 30 - H \\ C & = 30 - 21 = 9 \end{aligned}

In summary, Desi needs to mix

21

pounds of the Honduran coffee beans with

9

pounds of the Columbian coffee beans to create this blend.

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Subsection 4.2.3 Solving Special Systems of Equations with Substitution

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Remember the two special cases for a system of two linear equations? We studied them in Subsection 4.1.3. If the two lines have the same slope, then they might be distinct lines that never meet, and then the system has no solutions. Or they might coincide as the same line, in which case there are infinitely many solutions represented by all the points on that line. In these cases, when we try to use substitution, interesting things happen.

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Example 4.2.13. A System with No Solution.

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Solve the system of equations using the substitution method:

\begin{array}{r} {\begin{aligned} y & = 2 x - 1 \\ 4 x - 2 y & = 3 \end{aligned} \end{array}

Explanation.

Since the first equation has already isolated

y,

we will substitute

2 x - 1

in for

y

in the second equation, and we have:

\begin{aligned} 4 x - 2 y & = 3 \\ 4 x - 2 (2 x - 1) & = 3 \\ 4 x - 4 x + 2 & = 3 \\ 2 & = 3 \end{aligned}

Even though we were only intending to replace

y

in the second equation, it turned out that all instances of

x

disappear too. There are no variables at all in what remains. This will happen whenever the two lines from a system have the same slope. In this case, since

2 = 3

is false no matter what values

x

and

y

might be, there can be no solution to the system. If we graphed the two lines from this system, we would see them as parallel and distinct. We can say that there are no solutions, or that the solution set is empty. The solution set is

{},

\emptyset .

For verification, let’s re-write the second equation in slope-intercept form:

\begin{aligned} 4 x - 2 y & = 3 \\ - 2 y & = - 4 x + 3 \\ \frac{- 2 y}{- 2} & = \frac{- 4 x + 3}{- 2} \\ y & = \frac{- 4 x}{- 2} + \frac{3}{- 2} \\ y & = 2 x - \frac{3}{2} \end{aligned}

So the system is equivalent to:

\begin{array}{r} {\begin{aligned} y & = 2 x - 1 \\ y & = 2 x - \frac{3}{2} \end{aligned} \end{array}

Now it is easier to see that the two lines have the same slope but different

y

-intercepts. The lines are parallel and distinct confirming that there should be no solution to the system.

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Example 4.2.14. A System with Infinitely Many Solutions.

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Solve the system of equations using the substitution method:

\begin{array}{r} {\begin{aligned} y & = 2 x - 1 \\ 4 x - 2 y & = 2 \end{aligned} \end{array}

Explanation.

Since

y

is already isolated in the first line, we will use that and substitute

2 x - 1

in for

y

in the second equation:

\begin{aligned} 4 x - 2 y & = 2 \\ 4 x - 2 (2 x - 1) & = 2 \\ 4 x - 4 x + 2 & = 2 \\ 2 & = 2 \end{aligned}

Once again, after the substitution we find ourselves with an equation where there are no variables. This time, it is an outright true equation;

2

really does equal

2 .

What does this mean for the original system of equations though? Apparently as long as

y

equals

2 x - 1,

then both of the original equations are true. It’s obvious that

y = 2 x - 1

makes the first equation true, but our algebra above shows that

y = 2 x - 1

also makes the second equation true. So there are infinitely many solutions. If we want to, we can write the solution set using set-builder notation:

{(x, y) ∣ y = 2 x - 1} .

Ultimately the “two” lines from the system were actually the same line.

For verification, let’s re-write the second equation in slope-intercept form:

\begin{aligned} 4 x - 2 y & = 2 \\ - 2 y & = - 4 x + 2 \\ \frac{- 2 y}{- 2} & = \frac{- 4 x}{- 2} + \frac{2}{- 2} \\ y & = 2 x - 1 \end{aligned}

So the system is equivalent to:

\begin{array}{r} {\begin{aligned} y & = & 2 x - 1 \\ y & = & 2 x - 1 \end{aligned} \end{array}

Now it is clear that the two equations represent the same line. So every point on that line is a solution, and there are infinitely many solutions.

🔗

Reading Questions 4.2.4 Reading Questions

🔗

1.

🔗

Give an example of a system of two equations in

x

and

y

where it would be nicer to solve the system using substitution than by graphing the two lines that the equations define. Explain why substitution would be nicer than graphing for your example system.

🔗

2.

🔗

What can be helpful if you have a system of two linear equations in two variables where there are fractions appearing in the equations?

🔗

3.

🔗

In an application problem, thinking about the can help you understand how to set up equations.

🔗

Solve the system of equations using substitution.

🔗

1.

🔗

{\begin{aligned} D = - 4 E + 11 \\ D = 4 E - 21 \end{aligned}

🔗

2.

🔗

{\begin{aligned} l = - 5 K - 23 \\ l = - K + 1 \end{aligned}

🔗

3.

🔗

{\begin{aligned} V = - 5 (Q + 1) + 13 \\ V = 4 (Q + 3) - 13 \end{aligned}

🔗

4.

🔗

{\begin{aligned} B = - 5 (W + 1) + 37 \\ B = - (W - 2) + 2 \end{aligned}

🔗

5.

🔗

{\begin{aligned} - 5 c + k = 4 \\ c - 5 k = 28 \end{aligned}

🔗

6.

🔗

{\begin{aligned} - 5 h - R = - 29 \\ - h - R = - 9 \end{aligned}

🔗

7.

🔗

{\begin{aligned} A = 5 n + 22 \\ 4 n + 4 A = - 8 \end{aligned}

🔗

8.

🔗

{\begin{aligned} h = 5 t - 14 \\ - t - 3 h = 10 \end{aligned}

🔗

9.

🔗

{\begin{aligned} P = - 5 (z - 5) - 53 \\ P = 4 z + 35 \end{aligned}

🔗

10.

🔗

{\begin{aligned} x = 5 (E + 5) - 16 \\ x = - 2 E + 2 \end{aligned}

🔗

11.

🔗

{\begin{aligned} 8 J - f = 9 \\ 16 J - 2 f = - 7 \end{aligned}

🔗

12.

🔗

{\begin{aligned} - Q - 7 L = 9 \\ - 4 Q - 28 L = - 2 \end{aligned}

🔗

13.

🔗

{\begin{aligned} v = - 3 W + 18 \\ 3 W + v = 18 \end{aligned}

🔗

14.

🔗

{\begin{aligned} e = 2 c + 14 \\ - 2 c + e = 14 \end{aligned}

🔗

15.

🔗

{\begin{aligned} h - K = 3 \\ K = - 5 (h + 2) + 12 \end{aligned}

🔗

16.

🔗

{\begin{aligned} n + u = - 7 \\ u = 2 (n - 2) - 17 \end{aligned}

🔗

17.

🔗

{\begin{aligned} - 2 t + b = 5 \\ b = 2 (t - 4) + 16 \end{aligned}

🔗

18.

🔗

{\begin{aligned} 4 y + I = 0 \\ I = - 4 (y + 2) + 15 \end{aligned}

🔗

19.

🔗

{\begin{aligned} 2 E - 2 r = 2 \\ 5 E + 5 r = - 45 \end{aligned}

🔗

20.

🔗

{\begin{aligned} 4 J + 5 Z = 43 \\ - 3 J - 4 Z = - 34 \end{aligned}

🔗

21.

🔗

{\begin{aligned} F = 5 Q - 8 \\ F = 9 Q + 3 \end{aligned}

🔗

22.

🔗

{\begin{aligned} n = 5 W - 1 \\ n = 2 W - 3 \end{aligned}

🔗

23.

🔗

{\begin{aligned} 16 c + 4 X = 72 \\ X = - 4 (c + 3) + 30 \end{aligned}

🔗

24.

🔗

{\begin{aligned} - 12 h - 3 E = 21 \\ E = - 4 (h + 2) + 1 \end{aligned}

🔗

25.

🔗

{\begin{aligned} l = - 3 (m - 4) + 6 \\ l = 2 m + 5 \end{aligned}

🔗

26.

🔗

{\begin{aligned} V = - 2 (t + 7) + 5 \\ V = 2 (t + 3) - 4 \end{aligned}

🔗

27.

🔗

{\begin{aligned} 2 y - C = 9 \\ - 9 y + 7 C = 1 \end{aligned}

🔗

28.

🔗

{\begin{aligned} E - j = - 4 \\ 5 E - 4 j = 9 \end{aligned}

🔗

29.

🔗

{\begin{aligned} 5 J - 2 S = - 3 \\ S = - 2 J - 8 \end{aligned}

🔗

30.

🔗

{\begin{aligned} - 4 Q - z = - 1 \\ z = 2 Q - 3 \end{aligned}

🔗

31.

🔗

{\begin{aligned} h = - (W - 2) - 8 \\ 5 W - 2 h = 4 \end{aligned}

🔗

32.

🔗

{\begin{aligned} Q = - (b - 1) + 4 \\ - 9 b + 7 Q = 2 \end{aligned}

🔗

33.

🔗

{\begin{aligned} - 3 h + 2 y = 5 \\ 6 h - 3 y = - 6 \end{aligned}

🔗

34.

🔗

{\begin{aligned} - 2 m - 2 f = 1 \\ 8 m + 5 f = 6 \end{aligned}

🔗

35.

🔗

{\begin{aligned} N = 6.3 t + 2.6 \\ N = 4.6 t + 4.5 \end{aligned}

🔗

36.

🔗

{\begin{aligned} v = - 3.1 y + 2.4 \\ v = 0.1 y - 3.6 \end{aligned}

🔗

37.

🔗

{\begin{aligned} 5.8 E + 5.1 d = - 7.5 \\ - E + 6.2 d = - 7.1 \end{aligned}

🔗

38.

🔗

{\begin{aligned} 1.1 J - K = - 2.6 \\ 7.7 J + 2.2 K = 0.2 \end{aligned}

🔗

39.

🔗

{\begin{aligned} t = \frac{4}{3} P - \frac{1}{5} \\ t = \frac{5}{6} P + \frac{2}{3} \end{aligned}

🔗

40.

🔗

{\begin{aligned} b = \frac{5}{6} W - \frac{4}{5} \\ b = \frac{4}{5} W - \frac{2}{3} \end{aligned}

🔗

41.

🔗

{\begin{aligned} I = \frac{2}{5} (b + 3) - \frac{9}{5} \\ I = \frac{1}{3} (b - 4) + \frac{26}{15} \end{aligned}

🔗

42.

🔗

{\begin{aligned} s = \frac{5}{2} (h - 1) + \frac{19}{10} \\ s = \frac{5}{3} (h + 2) - \frac{23}{6} \end{aligned}

🔗

43.

🔗

{\begin{aligned} \frac{3}{5} m - \frac{5}{2} Z = \frac{1}{5} \\ \frac{2}{5} m + Z = - \frac{2}{5} \end{aligned}

🔗

44.

🔗

{\begin{aligned} \frac{1}{5} s - \frac{5}{3} G = - \frac{4}{3} \\ \frac{1}{6} s - G = \frac{4}{3} \end{aligned}

🔗

45.

🔗

{\begin{aligned} \frac{1}{2} y - \frac{1}{2} p = \frac{3}{4} \\ \frac{3}{4} y - \frac{3}{4} p = - \frac{1}{6} \end{aligned}

🔗

46.

🔗

{\begin{aligned} \frac{3}{2} D + \frac{3}{2} X = \frac{1}{4} \\ \frac{3}{4} D + \frac{3}{4} X = \frac{1}{4} \end{aligned}

🔗

47.

🔗

{\begin{aligned} \frac{2}{5} J - \frac{3}{2} D = \frac{4}{15} J + 1 \\ \frac{1}{3} J - D = \frac{5}{4} \end{aligned}

🔗

48.

🔗

{\begin{aligned} P + \frac{1}{5} m = - 5 P + 3 \\ \frac{5}{2} P + \frac{3}{5} m = \frac{1}{2} \end{aligned}

🔗

49.

🔗

{\begin{aligned} \frac{4}{5} W + T = - \frac{4}{5} (W + 3) - \frac{26}{15} \\ \frac{5}{4} W + \frac{3}{2} T = - \frac{4}{5} \end{aligned}

🔗

50.

🔗

{\begin{aligned} \frac{5}{6} b + \frac{5}{6} C = - (b + 1) - \frac{13}{10} \\ \frac{2}{3} b + C = \frac{5}{3} \end{aligned}

🔗

51.

🔗

{\begin{aligned} \frac{3}{4} g - \frac{5}{3} l = \frac{3}{2} \\ \frac{3}{2} g - \frac{10}{3} l = 3 \end{aligned}

🔗

52.

🔗

{\begin{aligned} \frac{5}{6} m - \frac{1}{4} S = \frac{1}{4} \\ \frac{3}{2} m - \frac{9}{20} S = \frac{9}{20} \end{aligned}

🔗

Applications

🔗

53.

🔗

A rectangle’s length is

8 f t

shorter than four times its width. The rectangle’s perimeter is

139 f t .

Find the rectangle’s length and width.

🔗

54.

🔗

A school fundraising event sold a total of

145

tickets and generated a total revenue of

$ 746.50 .

Each adult ticket cost

$ 6.50,

and each child ticket cost

$ 2.50 .

How many adult tickets and how many child tickets were sold?

🔗

55.

🔗

One telecom company charges a monthly fee of

$ 36.95

and

$ 4.25

for each Gigabyte (GB) of data transmitted. A rival telecom company charges a monthly fee of

$ 45.95

and

$ 3.20

for each GB of data transmitted. How many GB of data would you have to use for the monthly bill to be the same for the two providers? And what would that monthly bill be?

🔗

56.

🔗

A local restaurant has two locations. At one location, the revenue this month is

$ 94,000

but it has been decreasing by

$ 5,000

per month. At the other location, the annual revenue this month is

$ 49,000

and it has been increasing by

$ 2,500

per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be?

🔗

57.

🔗

An algebra exam has

29

questions worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth

2

points, and short-answer questions, each worth

8

points. How many questions are there of each type?

🔗

58.

🔗

Skye invested a total of

$ 4,200

in two investments. Her savings account pays

2.5 %

interest annually. A riskier stock investment earned

5 %

at the end of the year. At the end of the year, Skye earned a total of

$ 175

in interest. How much money did she invest in each account?

🔗

59.

🔗

Alessandro invested a total of

$ 3,900

in two investments. His savings account pays

1.5 %

interest annually. A riskier stock investment lost

7 %

at the end of the year. At the end of the year, Alessandro’s total fell from

$ 3,900

$ 3,687.50 .

How much money did he invest in each account?

🔗

60.

🔗

Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with

46 %

Democrats. Balsamburg had a population with

51 %

Democrats. After the merge, the new city has a total of

3700

residents, with

49.6 %

Democrats. How many residents did each town have before the merge?

🔗

61.

🔗

Delilah poured some

10 %

alcohol solution and some

18 %

alcohol solution together into a beaker, and then the beaker had

555 m l

12.6 %

alcohol solution. How much of each solution did Delilah pour into the beaker?

🔗

62.

🔗

A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost

$ 1.67

per pound, and cashews cost

$ 5.75

per pound. The company is targeting a product that will cost them

$ 2.50

per pound worth of ingredients. How much of each type of nut should go into a bag?

You have attempted 1 of 68 activities on this page.

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Section 4.2 Substitution

Objectives: PCC Course Content and Outcome Guide

Subsection 4.2.1 Substitution

Example 4.2.2. The Interview.

Remark 4.2.3.

Example 4.2.4.

Checkpoint 4.2.5.

(a)

(b)

(c)

(d)

Example 4.2.6.

Example 4.2.7.

Example 4.2.8.

Process 4.2.9. Solving Systems of Equations by Substitution.

Subsection 4.2.2 Applications

Example 4.2.10. Two Different Interest Rates.

Example 4.2.11. Mixing Solutions with Two Different Concentrations.

Checkpoint 4.2.12. Mixing a Coffee Blend.

(a)

(b)

(c)

(d)

Subsection 4.2.3 Solving Special Systems of Equations with Substitution

Example 4.2.13. A System with No Solution.

Example 4.2.14. A System with Infinitely Many Solutions.

Reading Questions 4.2.4 Reading Questions

1.

2.

3.

Exercises 4.2.5 Exercises

Skills Practice

Exercise Group.

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