Systems of Equations

(12, 16)

Multiply the first equation by 2: $4x + 2y = 80$. Add it to $x - 2y = -20$: $5x = 60$, so $x = 12$. Substitute: $2(12) + y = 40$, so $y = 16$. Check: $12 - 32 = -20$. ✓

(15, 17.5): A coefficient was switched when setting up elimination. Make the y-coefficients opposites by multiplying the whole first equation by 2.

There is no solution.: Elimination was set up incorrectly. These lines have different slopes, so they cross at exactly one point.

There are an infinite number of solutions.: The equations aren't multiples of each other, so the lines are not the same line. They meet at one point.

5.25 mi

Let $m$ = morning miles and $a$ = afternoon miles: $16m + 12a = 141$ and $10m + 15a = 123.75$. Multiply the first by 1.25: $20m + 15a = 176.25$. Subtract the second: $10m = 52.5$, so $m = 5.25$.

6.00 mi: This is $16 - 10$, the change in morning trips, not a distance. Write and solve a system of two equations.

4.75 mi: The system was solved correctly, but this is the afternoon route. The question asks for the morning route.

5.00 mi: Check the arithmetic. Substitute your values into both equations to confirm they give 141 and 123.75.

13

Let $t$ = 3-credit courses and $f$ = 4-credit courses: $t + f = 18$ and $3t + 4f = 59$. Substitute $f = 18 - t$: $3t + 72 - 4t = 59$, so $-t = -13$ and $t = 13$.

5: That's the number of 4-credit courses. Re-read which variable the question asks for.

20: This is the credits earned from the 4-credit courses ($4 \cdot 5$), not a number of courses.

39: This is the credits from the 3-credit courses ($3 \cdot 13$). The question asks how many courses.

6.5

Substitute $y = -5x + 32$ into the first equation: $3x - 5(-5x + 32) = 22$, so $28x - 160 = 22$. Add 160: $28x = 182$, so $x = 6.5$.

−6.5: A sign error happened when moving 160. Add 160 to both sides: $28x = 182$.

0.5: This comes from sign errors and finding y instead of x. The question asks for x.

−0.5: This is the y-value of the solution. The question asks for x.

−17

Substitute $x=-2y+9$: $6y-2y+9=-59$, so $4y=-68$ and $y=-17$.

43: This comes from combining terms or signs incorrectly; $4y=-68$.

8.5: Check the substitution: $6y-2y=4y$, and $-59-9=-68$.

−12.5: Use $6y-2y=4y$, not $8y$, before dividing.

10

Let $x$ be children’s shirts and $y$ be adults’ shirts. Then $x + y = 14$ and $15x + 19y = 250$. Solving gives $y = 10$.

4: This results from transposing the 15 and 19 coefficients when setting up the system.

7: This assumes the same number of children’s and adults’ shirts were sold. The two prices require a system of equations.

14: This is the total number of shirts, not the number of adults’ shirts.

$\frac{2}{3}$

Substitute $y = -6x + 2$: $3x - 5(-6x + 2) = 12$, so $3x + 30x - 10 = 12$. Then $33x = 22$, and $x = \frac{22}{33} = \frac{2}{3}$.

$\frac{22}{3}$: Only the last term in the parentheses was multiplied by −5, and the $x$-terms were combined with a sign error, leading to $3x = 22$.

$-\frac{10}{3}$: The −5 was not distributed, giving $3x - 6x + 2 = 12$, or $-3x = 10$.

$-\frac{18}{3}$: The constant in the parentheses was ignored and $30x$ was subtracted from 12 without its variable, giving $3x = -18$.

The system has infinitely many solutions.

Substitute $y = -0.5x + 4$: $3x + 6(-0.5x + 4) = 24$, so $3x - 3x + 24 = 24$, or $24 = 24$. The statement is always true, so both equations describe the same line and the system has infinitely many solutions.

The system has no solutions.: The $x$-terms were eliminated, but $24 = 24$ is always true, which means infinitely many solutions, not none.

The system has exactly one solution.: A point such as $(0, 4)$ satisfies both equations, but so does every other point on the line.

The system has exactly two solutions.: Points such as $(0, 4)$ and $(8, 0)$ satisfy both equations, but two lines can't intersect at exactly two points.

$\frac{1}{3}$

Substitute $y = x - 1$: $6x + 3(x - 1) = 0$, so $9x - 3 = 0$. Then $9x = 3$ and $x = \frac{1}{3}$.

$-\frac{1}{3}$: 3 was subtracted instead of added when solving $9x - 3 = 0$.

$-\frac{2}{3}$: This is the $y$-value of the solution, not the $x$-value.

$\frac{2}{3}$: This is the $y$-value with the subtraction reversed: $1 - \frac{1}{3}$.