Unit 4 · Checkpoint quiz

Unit 4 checkpoint: Systems of linear equations

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next.

Which point is a solution of the system $x + 2y = 4$ and $3x - y = 5$?

$(4, 0)$
$(0, 2)$
$(2, 1)$
$(1, -2)$

A solution must make both equations true. Test each point in both.

(a) $4 + 0 = 4$ works, but $3(4) - 0 = 12$, not 5. (b) $0 + 4 = 4$ works, but $0 - 2 = -2$, not 5.

(d) $1 + 2(-2) = -3$, not 4. Answer: (c) $(2, 1)$.

Solve by substitution: $y = 2x - 5$ and $3x + 2y = 11$. Give the answer as a point $(x, y)$.

Replace $y$ in the second equation with $2x - 5$: $3x + 2(2x - 5) = 11$.

Distribute and combine: $3x + 4x - 10 = 11$, so $7x = 21$ and $x = 3$.

Find $y$: $y = 2(3) - 5 = 1$. Check: $3(3) + 2(1) = 11$. Answer: $(3, 1)$.

Solve by substitution: $x - 2y = 8$ and $3x + 4y = 4$.

Solve the first equation for $x$: $x = 8 + 2y$.

Substitute: $3(8 + 2y) + 4y = 4$, so $24 + 6y + 4y = 4$, $10y = -20$, $y = -2$.

Find $x$: $x = 8 + 2(-2) = 4$. Check: $3(4) + 4(-2) = 4$. Answer: $(4, -2)$.

Solve by elimination: $x + 3y = 5$ and $2x - y = -4$.

Multiply the second equation by 3 so the $y$ terms are opposites: $6x - 3y = -12$.

Add it to the first equation: $(x + 6x) + (3y - 3y) = 5 + (-12)$, so $7x = -7$ and $x = -1$.

Find $y$: $-1 + 3y = 5$, so $3y = 6$, $y = 2$. Check: $2(-1) - 2 = -4$. Answer: $(-1, 2)$.

Solve by elimination: $2x + 3y = -1$ and $5x - 4y = 32$.

Make the $y$ terms opposites: multiply the first equation by 4 and the second by 3.

$8x + 12y = -4$ and $15x - 12y = 96$. Add: $23x = 92$, so $x = 4$.

Find $y$: $2(4) + 3y = -1$, so $3y = -9$, $y = -3$. Check: $5(4) - 4(-3) = 20 + 12 = 32$. Answer: $(4, -3)$.

Solve the system $2x - 4y = 6$ and $-x + 2y = -3$. If there is no solution type none; if there are infinitely many, type infinite.

Multiply the second equation by 2: $-2x + 4y = -6$.

Add it to the first: $0x + 0y = 0$, which is $0 = 0$. This is always true.

The two equations describe the same line, so every point on it is a solution. Answer: infinitely many solutions.

Without solving, what does the graph of the system $y = 3x + 1$ and $6x - 2y = 8$ look like?

Parallel lines, no solution
Crossing lines, one solution
Same line, infinitely many solutions
Perpendicular lines, one solution

Write the second equation in slope-intercept form: $-2y = -6x + 8$, so $y = 3x - 4$.

Both lines have slope 3, but the y-intercepts are different (1 and $-4$).

Same slope, different intercepts means parallel lines that never meet. Answer: (a) no solution.

A school play sold 40 tickets for a total of $296. Adult tickets cost $9 and child tickets cost $5. How many adult tickets were sold?

Let $a$ = adult tickets and $c$ = child tickets. Then $a + c = 40$ and $9a + 5c = 296$.

From the first equation, $c = 40 - a$. Substitute: $9a + 5(40 - a) = 296$, so $9a + 200 - 5a = 296$.

$4a = 96$, so $a = 24$ (and $c = 16$). Check: $9(24) + 5(16) = 216 + 80 = 296$. Answer: 24 adult tickets.

A boat goes 36 miles downstream (with the current) in 2 hours. The return trip upstream (against the current) takes 3 hours. What is the speed of the boat in still water, in miles per hour?

Let $b$ = boat speed and $c$ = current speed. Downstream rate: $36 \div 2 = 18$, so $b + c = 18$.

Upstream rate: $36 \div 3 = 12$, so $b - c = 12$.

Add the equations: $2b = 30$, so $b = 15$ (and $c = 3$). Answer: 15 miles per hour.

Solve: $x + y + z = 6$, $2x - y + z = 3$, and $x + 2y - z = 2$. Give the answer as $(x, y, z)$.

Add the first and third equations to eliminate $z$: $2x + 3y = 8$.

Add the second and third equations to eliminate $z$: $3x + y = 5$.

From $3x + y = 5$, $y = 5 - 3x$. Substitute: $2x + 3(5 - 3x) = 8$, so $-7x + 15 = 8$, $x = 1$, and $y = 5 - 3 = 2$.

Find $z$ from the first equation: $1 + 2 + z = 6$, so $z = 3$. Answer: $(1, 2, 3)$.

Write the equation of the line through $(2, -1)$ that is parallel to $y = 3x + 4$. Give it in the form $y = mx + b$.

Parallel lines have the same slope, so $m = 3$.

Point-slope form: $y - (-1) = 3(x - 2)$, so $y + 1 = 3x - 6$.

Subtract 1: $y = 3x - 7$.

Solve $-3x + 5 \le 14$. Give the answer in interval notation.

Subtract 5: $-3x \le 9$.

Divide by $-3$ and flip the inequality sign: $x \ge -3$.

In interval notation: $[-3, \infty)$. The bracket means $-3$ is included.

Want to look something up first? Lines and systems cheat sheet · Glossary

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