Word Problems with Linear Equations: Practice

Let $n$ be the smaller number, so the larger is $2n + 6$. Their sum is $54$.

$n + (2n + 6) = 54, \qquad 3n + 6 = 54, \qquad n = 16$

The smaller is $16$ and the larger is $2(16) + 6 = 38$; check $16 + 38 = 54$.

Let $n$ be the smallest, so the even integers are $n$, $n + 2$, $n + 4$ with sum $3n + 6$. This sum is $2$ less than $5n$.

$3n + 6 = 5n - 2, \qquad 8 = 2n, \qquad n = 4$

The integers are $4$, $6$, $8$ with sum $18$, and $5(4) - 2 = 18$, so the smallest is $4$.

Let $t$ be the time in hours. Going the same way, the gap grows at the difference of their speeds, $5 - 3 = 2$ mph.

$(5 - 3)t = 6, \qquad 2t = 6, \qquad t = 3$

Check: in $3$ hours they walk $15$ and $9$ miles, and $15 - 9 = 6$, so they are $6$ miles apart after $3$ hours.

Let $x$ be the first angle in degrees, so the others are $2x$ and $3x$. A triangle's angles sum to $180$.

$x + 2x + 3x = 180, \qquad 6x = 180, \qquad x = 30$

The angles are $30$, $60$, and $90$ degrees; check $30 + 60 + 90 = 180$, so the first is $30$ degrees.

There are $4$ more nickels than dimes, so the nickels number $d + 4$. Nickels are worth $5$ cents each and dimes $10$ cents each.

$5(d + 4) + 10d = 200$

The nickels carry the $d + 4$, not the dimes, so the value is $5(d + 4) + 10d$.

Let $n$ be the number. Half of it is $\frac{1}{2}n$ and a third is $\frac{1}{3}n$, with the half being $5$ larger. Multiply through by $6$ to clear the fractions.

$\tfrac{1}{2}n = \tfrac{1}{3}n + 5, \qquad 3n = 2n + 30, \qquad n = 30$

Check: half of $30$ is $15$ and a third is $10$, and $15 = 10 + 5$, so the number is $30$.

Let $p$ be the pounds of peanuts, so the cashews weigh $15 - p$ pounds. The value of peanuts plus cashews equals the value of the mix, $6 \times 15 = 90$ dollars.

$4p + 9(15 - p) = 90, \qquad -5p + 135 = 90, \qquad p = 9$

Use $9$ pounds of peanuts ($36$ dollars) and $6$ pounds of cashews ($54$ dollars); check $36 + 54 = 90$.

Let $n$ be the first, so the integers are $n$, $n + 1$, $n + 2$. The first and third add to $34$.

$n + (n + 2) = 34, \qquad 2n + 2 = 34, \qquad n = 16$

The integers are $16$, $17$, $18$, so the middle one is $17$; check $16 + 18 = 34$.

Let $b$ be Ben's age, so Ann is $b + 8$. Their ages sum to $40$.

$b + (b + 8) = 40, \qquad 2b + 8 = 40, \qquad b = 16$

Ben is $16$ and Ann is $24$; check $16 + 24 = 40$ and $24 - 16 = 8$.

Increasing the number by $25\%$ of itself adds $0.25n$ to $n$, and the result is $60$.

$n + 0.25n = 60$

This is $1.25n = 60$, giving $n = 48$, so the model is $n + 0.25n = 60$.

Let $q$ be the number of quarters, so the dimes number $2q$. In cents, the total value is $900$.

$25q + 10(2q) = 900, \qquad 45q = 900, \qquad q = 20$

There are $20$ quarters ($500$ cents) and $40$ dimes ($400$ cents); check $500 + 400 = 900$ cents, or $9$ dollars.

Consecutive multiples of $5$ differ by $5$, so let the smaller be $n$ and the larger $n + 5$. Their sum is $65$.

$n + (n + 5) = 65, \qquad 2n + 5 = 65, \qquad n = 30$

The multiples are $30$ and $35$, so the smaller is $30$; check $30 + 35 = 65$.