Word Problems with Linear Equations: Practice

Let $n$ be the smaller number, so the larger is $n + 8$. Their sum is $50$.

$n + (n + 8) = 50, \qquad 2n + 8 = 50, \qquad n = 21$

The smaller is $21$, so the larger is $21 + 8 = 29$; check $21 + 29 = 50$.

Let $n$ be the smallest. Consecutive odd integers step by $2$, so they are $n$, $n + 2$, and $n + 4$.

$n + (n + 2) + (n + 4) = 63, \qquad 3n + 6 = 63, \qquad n = 19$

The integers are $19$, $21$, and $23$, so the largest is $23$; check $19 + 21 + 23 = 63$.

Let $w$ be the width in centimeters, so the length is $w + 5$. The perimeter is twice the sum of length and width.

$2(w + (w + 5)) = 46, \qquad 4w + 10 = 46, \qquad w = 9$

The width is $9$ cm and the length is $14$ cm; check $2(9 + 14) = 46$ cm.

In $x$ years each is $x$ years older, so Maria is $30 + x$ and her son is $6 + x$. Maria being twice as old means her age equals $2$ times the son's age then.

$30 + x = 2(6 + x)$

Both ages shift by the same $x$, which is why the son's future age is $6 + x$ inside the doubling.

Let $t$ be the time in hours. In $t$ hours the trains cover $40t$ and $50t$ miles, and moving apart, the gap is the sum.

$40t + 50t = 270, \qquad 90t = 270, \qquad t = 3$

Check: in $3$ hours they go $120$ and $150$ miles, and $120 + 150 = 270$, so the time is $3$ hours.

Let $n$ be the smallest, so the integers are $n$, $n + 1$, and $n + 2$. Their sum is $72$.

$n + (n + 1) + (n + 2) = 72, \qquad 3n + 3 = 72, \qquad n = 23$

The integers are $23$, $24$, and $25$, so the middle one is $24$.

Each quarter is worth $25$ cents, so $x$ quarters are worth $25x$ cents. The $8$ dimes are worth $8 \times 10 = 80$ cents.

$25x + 80$

Adding the two values gives the total, $25x + 80$ cents.

The number decreased by $9$ is $n - 9$, and $4$ times the number is $4n$. The words the same result as make these equal.

$n - 9 = 4n$

Decreasing $n$ means $n - 9$, not $9 - n$, so the correct model is $n - 9 = 4n$.

Let $t$ be the total number of questions. Getting $90\%$ of them right means $0.90t = 27$.

$0.90t = 27, \qquad t = \frac{27}{0.90} = 30$

Check: $90\%$ of $30$ is $27$, so the test had $30$ questions.

Let $n$ be the number of nickels, so the dimes number $n + 3$. In cents, nickels are worth $5n$ and dimes $10(n + 3)$.

$5n + 10(n + 3) = 195, \qquad 15n + 30 = 195, \qquad n = 11$

There are $11$ nickels ($55$ cents) and $14$ dimes ($140$ cents); check $55 + 140 = 195$.

Five years ago Ben was $5$ years younger than he is now, so his age then was $b - 5$, and that equals $12$.

$b - 5 = 12$

Solving gives $b = 17$, so the model is $b - 5 = 12$.

Let $w$ be the width in centimeters, so the length is $3w - 1$. The perimeter is twice the sum of length and width.

$2(w + (3w - 1)) = 38, \qquad 8w - 2 = 38, \qquad w = 5$

The width is $5$ cm and the length is $14$ cm; check $2(5 + 14) = 38$ cm.