Systems of Equations — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

Three customers report what they paid at a food truck for different orders. From those three receipts you can work out the price of each item. Finding several unknowns from several facts is solving a system, and it is how every model in this course gets its constants.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the slope of $y = -2x + 7$ ?

$-2$
$7$
$2$
$-\frac{1}{2}$

Show the answer

In $y = mx + b$ , the slope is $m = -2$ .

Correct: $-2$ :
$7$ :
$2$ :
$-\frac{1}{2}$ :

2. Does the point $(1, 5)$ satisfy $y = x^2 + 4$ ?

Yes
No, it gives $y = 6$
No, it gives $y = 4$
No, it gives $y = 3$

Show the answer

$1^2 + 4 = 5$ , so the point is on the graph.

Correct: Yes:
No, it gives $y = 6$ :
No, it gives $y = 4$ :
No, it gives $y = 3$ :

Part 4 · See it

See it first

The lines y = 2x - 1 and y = -x + 5 crossing at the point (2, 3), the solution of the system. — The system $y = 2x - 1$ and $y = -x + 5$ has one solution, $(2, 3)$ , the point where the lines cross. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Write one equation for each fact (each data point)a system with as many equations as unknowns
Substitute or add equations to remove one variable at a timeone equation in one variable, which you solve
Work back up and check every original equationall the unknowns, at full precision

Part 6 · Key ideas

Key ideas

A solution of a system makes every equation true at once; for two lines it is where they cross.
To fit $y = ax^2 + bx + c$ to three points, substitute each point to get three equations in $a$ , $b$ and $c$ .
With a calculator, store each constant; rounding early spoils every value that comes after.

Part 7 · Misconception

A common mistake

The wrong idea: Substituting $(2, 11)$ into $y = ax^2 + bx + c$ gives $2a + 2b + c = 11$ .

What actually happens: The input is squared for the $a$ term: $a(2)^2 + 2b + c = 11$ , so $4a + 2b + c = 11$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Solve the system $y = x - 1$ and $3x + 2y = 13$ by substitution. What is $x$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Substitute $y = x - 1$ : $3x + 2(x - 1) = 13$ , so $5x - 2 = 13$ , $5x = 15$ and $x = 3$ . Then $y = 3 - 1 = 2$ . Check: $3(3) + 2(2) = 13$ .

Answer: 3

2. Solve $2x + 3y = 12$ and $4x - 3y = 6$ by elimination. What is $y$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Add the equations to eliminate $y$ : $6x = 18$ , so $x = 3$ . Substitute: $2(3) + 3y = 12$ , so $3y = 6$ and $y = 2$ .

Answer: 2

3. A quadratic $y = ax^2 + bx + c$ passes through $(1, 6)$ , $(2, 11)$ and $(3, 18)$ . Which system gives $a$ , $b$ and $c$ ?

$a + b + c = 6$ , $4a + 2b + c = 11$ , $9a + 3b + c = 18$
$a + b + c = 6$ , $2a + 2b + c = 11$ , $3a + 3b + c = 18$
$6a + 6b + c = 1$ , $11a + 11b + c = 2$ , $18a + 18b + c = 3$
$a + b + c = 6$ , $4a + 2b + c = 11$ , $6a + 3b + c = 18$

Show the answer

Substitute each point into $y = ax^2 + bx + c$ : at $(2, 11)$ , $a(2)^2 + b(2) + c = 11$ , which is $4a + 2b + c = 11$ ; the same for the others.

Correct: $a + b + c = 6$ , $4a + 2b + c = 11$ , $9a + 3b + c = 18$ : Each input is squared for the $a$ term.
$a + b + c = 6$ , $2a + 2b + c = 11$ , $3a + 3b + c = 18$ : The $x^2$ was not squared: $2^2 = 4$ , not $2$ .
$6a + 6b + c = 1$ , $11a + 11b + c = 2$ , $18a + 18b + c = 3$ : The inputs and outputs are swapped.
$a + b + c = 6$ , $4a + 2b + c = 11$ , $6a + 3b + c = 18$ : $3^2 = 9$ , not $6$ .

4. Solve the system $a + b + c = 6$ , $4a + 2b + c = 11$ , $9a + 3b + c = 18$ . What is $a$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Subtract the first equation from the second: $3a + b = 5$ . Subtract the second from the third: $5a + b = 7$ . Subtract these: $2a = 2$ , so $a = 1$ . Then $b = 5 - 3 = 2$ and $c = 6 - 1 - 2 = 3$ .

Answer: 1

5. What can you say about the system $y = 3x + 1$ and $6x - 2y = 4$ ?

It has one solution, at $(1, 4)$
It has many solutions, because the lines are the same
It has one solution, at $(0, 1)$
It has no solution, because the lines are parallel and different

Show the answer

Rewrite the second: $2y = 6x - 4$ , so $y = 3x - 2$ . Both lines have slope $3$ but different $y$ -intercepts ( $1$ and $-2$ ), so they never meet.

It has one solution, at $(1, 4)$ : $(1, 4)$ is on the first line, but $6 - 8 = -2 \ne 4$ .
It has many solutions, because the lines are the same: The intercepts differ, so the lines are not the same.
It has one solution, at $(0, 1)$ : $(0, 1)$ gives $0 - 2 = -2 \ne 4$ in the second equation.
Correct: It has no solution, because the lines are parallel and different: Equal slopes, different intercepts.

Part 9 · Summary

Summary

Solve a system by substitution or elimination, removing one variable at a time. Fitting a model through data points means writing one equation per point and solving for the constants, stored at full precision.

Part 10 · Up next

What comes next

Part 11 · Connections