Precalculus by Richard Wright

So we say with confidence, “The Lord is my helper; I will not be afraid. What can mere mortals do to me?” Hebrews 13:6 NIV

Summary: In this section, you will:

- Solve a system of equations by graphing.
- Solve a system of equations by substitution.

SDA NAD Content Standards (2018): PC.6.1

A ski jumper's path through the air is approximately parabolic with an equation in the form of \(y = ax^2 + bx +c\). The ground where he lands is sloped and is approximately linear until the bottom with an equation in the form of \(y = mx + b\). The point where he touches the ground is common to both equations. This lesson is about how to find any points in common to two equations.

A system of equations is more than one equation with a shared solution. A solution to a system of equations are all the points common to both graphs. The most basic way to solve a system is to graph both equations on the same graph and then look for any intersections. These points are the solutions.

- Graph both equations on the same graph.
- Any points of intersection are the solutions.

Solve \(\left\{\begin{align} y &= 2x - 3 \\ y &= x + 3 \end{align}\right.\) by graphing.

Graph both equations on the same graph. The solutions are where they intersect. In this case, there is one solution and it is (2, 1).

Solve \(\left\{\begin{align} x^2 - y = 4 \\ x^2 - 2x + y = 0 \end{align}\right.\) by graphing.

Graph both equations on the same graph. The solutions are where they intersect. In this case, there are two solutions, and they are (−1, −3) and (2, 0).

Solve \(\left\{\begin{align} y &= \frac{1}{2} x - 2 \\ y &= -2x + 3 \end{align}\right.\) by graphing.

(2, −1)

While graphing gives a nice picture of the solutions, often it requires estimating the solutions. A better way that works for almost all cases is substitution. For substitution, solve one equation for one variable. Then substitute that expression into the other equation. This can then be solved for the remaining variable. Finally, take that answer and substitute it into the first equation to get the final solution.

- Solve one equation for one variable.
- Substitute this expression into the other equation.
- Solve the new equation.
- Substitute the solution back into the first equation and solve.
- Check the solution.

Solve \(\left\{\begin{align} 2x + y &= 4 \\ -x + y &= 2 \end{align}\right.\) by substitution.

Solve one of the equations for a variable. Maybe start by solving the first equation for *y*.

$$ 2x + y = 4 $$

$$ y = -2x + 4 $$

Substitute this into the other equation.

$$ -x + y = 2 $$

$$ -x + \left(\color{blue}{-2x + 4}\right) = 2 $$

$$ -3x + 4 = 2 $$

$$ -3x = -2 $$

$$ x = \frac{2}{3} $$

Substitute this answer back into the first equation. It is easiest to substitute it into the solved version of the first equation.

$$ y = -2x + 4 $$

$$ y = -2\left(\frac{2}{3}\right) + 4 $$

$$ y = \frac{8}{3} $$

The solution is \(\left(\frac{2}{3}, \frac{8}{3}\right)\).

Solve \(\left\{\begin{align} \left(x - 1\right)^2 + \left(y + 1\right)^2 = 9 \\ x - y = -1 \end{align}\right.\) by substitution.

It is easiest to solve the second equation for a variable such as *x*.

$$ x - y = -1 $$

$$ x = y - 1 $$

Substitute this expression for *x* in the other equation.

$$ \left(x - 1\right)^2 + \left(y + 1\right)^2 = 9 $$

$$ \left(\left(\color{blue}{y - 1}\right) - 1\right)^2 + \left(y + 1\right)^2 = 9 $$

$$ \left(y - 2\right)^2 + \left(y + 1\right)^2 = 9 $$

$$ y^2 - 4y + 4 + y^2 + 2y + 1 = 9 $$

$$ 2y^2 - 2y - 4 = 0 $$

$$ 2\left(y^2 - y - 2\right) = 0 $$

$$ 2\left(y - 2\right)\left(y + 1\right) = 0 $$

\(y - 2 = 0\) and \(y + 1 = 0\)

\(y = 2\) and \(y = -1\)

Substitute both of these into the original equation.

$$ x = y - 1 $$

Use *y* = 2

$$ x = \color{blue}{2} - 1 = 1 $$

So one solution is (1, 2).

Use *y* = −1

$$ x = \color{blue}{-1} - 1 = -2 $$

So the other solution is (−2, −1).

The solutions are (1, 2) and (−2, −1).

Solve \(\left\{\begin{align} y = x^2 - 4 \\ 2x - y = 1 \end{align}\right.\)

(−1, −3) and (3, 5)

- Graph both equations on the same graph.
- Any points of intersection are the solutions.

- Solve one equation for one variable.
- Substitute this expression into the other equation.
- Solve the new equation.
- Substitute the solution back into the first equation and solve.
- Check the solution.

Helpful videos about this lesson.

- Find out how to solve by graphing on your graphing utility. Write the steps.
- \(\left\{\begin{align} 2x + y &= 3 \\ x - 3y &= 5 \end{align}\right.\)
- \(\left\{\begin{align} x - 2y &= -1 \\ 3x + 4y &= -3 \end{align}\right.\)
- \(\left\{\begin{align} 3x + y &= 1 \\ 2x + 2y &= 6 \end{align}\right.\)
- \(\left\{\begin{align} x^2 + y &= 3 \\ x + y &= 3 \end{align}\right.\)
- \(\left\{\begin{align} 2x - y^2 &= 0 \\ x^2 + y^2 &= 8 \end{align}\right.\)
- \(\left\{\begin{align} y &= x^2 - 4 \\ y &= -x^2 - 2 \end{align}\right.\)
- \(\left\{\begin{align} 3x + y &= 7 \\ 4x - 5y &= 22 \end{align}\right.\)
- \(\left\{\begin{align} 3x + 2y &= 7 \\ x + 4y &= 19 \end{align}\right.\)
- \(\left\{\begin{align} 3x - 5y &= -6 \\ x - y &= -1 \end{align}\right.\)
- \(\left\{\begin{align} 5x + y &= -5 \\ -7x + 3y &= -3 \end{align}\right.\)
- \(\left\{\begin{align} y &= 2x^2 \\ y &= -3x + 5 \end{align}\right.\)
- \(\left\{\begin{align} x^2 - 3y &= -2 \\ 2x + y &= -1 \end{align}\right.\)
- \(\left\{\begin{align} \frac{x^2}{9} + \frac{y^2}{4} &= 1 \\ 2x + 3y &= 6 \end{align}\right.\)
- A ski jumper leaves the end of the ski jump headed upward along a parabolic trajectory that can be modeled by \(y = -\frac{1}{10} x^2 + 2x\) in meters. The ski slope falls away in a straight line 45° below the horizontal that can be modeled by \(y = -x\). Measuring down the slope, how far from the end of the jump does the skier land on the surface?
- (7-09) Write the polar equation of a parabola with directrix
*y*= 10. - (6-03) Given \(\overset{\rightharpoonup}{p} = \langle 1, 3 \rangle\) and \(\overset{\rightharpoonup}{q} = \langle -2, 2 \rangle\), find \(2\overset{\rightharpoonup}{p} - \overset{\rightharpoonup}{q}\).
- (5-04) Find all the solutions of \(2 \sin 2x - \sqrt{3} = 0\).
- (4-02) Evaluate \(\tan \frac{11π}{6}\) without using a calculator.
- (3-03) Use the change-of-base formula and a calculator to evaluate log
_{3}28.

Solve by graphing.

Solve by substitution.

Problem Solving

Mixed Review

- Answers will vary but probably includes "intersect" command.
- (2, −1)
- (−1, 0)
- (−1, 4)
- (0, 3), (1, 2)
- (2, 2), (2, −2)
- (−1, −3), (1, −3)
- (3, −2)
- (−1, 5)
- \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
- \(\left(-\frac{6}{11}, -\frac{25}{11}\right)\)
- \(\left(-\frac{5}{2}, \frac{25}{2}\right)\), (1, 2)
- (−5, 9), (−1, 1)
- (0, 2), (3, 0)
*x*= 30,*y*= −30. Use Pythagorean theorem to find distance, \(30\sqrt{2} \approx 42.4 \text{ m}\).- \(r = \frac{10}{1 + \sin θ}\)
- \(\langle 4, 4 \rangle\)
- \(x = \frac{π}{6} + πn, \frac{π}{3} + πn\)
- \(-\frac{\sqrt{3}}{3}\)
- 3.033