We can approximate a linear pattern in a scatterplot using a regression line.
We can use interpolation or extrapolation to make estimates and predictions.
If we extrapolate too far beyond the known data, we may get unreasonable results.
A solution to a linear system is an ordered pair that satisfies both equations.
A solution to a linear system is a point where the two graphs intersect.
The graphs of the equations in an inconsistent system are parallel lines and hence to do not intersect.
The graphs of the two equations in a dependent system are the same line.
If a company’s revenue exactly equals its costs (so that their profit is zero), we say that the business venture will break even.
For solving a linear system, the substitution method is easier if one of the variables in one of the equations has a coefficient of or
If a linear combination of the equations in a system results in an equation of the form
then the system is inconsistent. If an equation of the form
results, then the system is dependent
The point-slope form is useful when we know the rate of change and one point on the line.
The solutions of the linear inequality
consists of the line and a half-plane on one side of that line.
To Graph an Inequality Using a Test Point.
Graph the corresponding equation to obtain the boundary line.
Choose a test point that does not lie on the boundary line.
Substitute the coordinates of the test point into the inequality.
If the resulting statement is true, shade the half-plane that includes the test point.
If the resulting statement is false, shade the half-plane that does not include the test point.
If the inequality is strict, make the boundary line a dashed line.
The solutions to a system of inequalities includes all points that are solutions to all the inequalities in the system. The graph of the system is the intersection of the shaded regions for each inequality in the system.
A solution to an equation in three variables is an ordered triple of numbers that satisfies the equation.
To solve a linear system, we use linear combinations to reduce the system to triangular form, and then use back-substitution to find the solutions.