Word Problems with Systems: Practice

Let $x$ and $y$ be the amounts at $5\%$ and $8\%$: $x + y = 2000$ and $0.05x + 0.08y = 139$. Substitute $x = 2000 - y$.

$0.05(2000 - y) + 0.08y = 139 \;\Rightarrow\; 0.03y = 39 \;\Rightarrow\; y = 1300$

So $1300$ dollars is invested at $8\%$ (and $700$ at $5\%$).

Let $t$ and $u$ be the tens and units digits: $t + u = 11$ and $t = u + 3$. Substitute $u + 3$ for $t$.

$(u + 3) + u = 11 \;\Rightarrow\; 2u = 8 \;\Rightarrow\; u = 4$

Then $t = 7$, so the number is $74$.

Let $x$ and $y$ be the pounds of the $12$- and $7$-dollar coffees: $x + y = 20$ and $12x + 7y = 9(20) = 180$. Substitute $y = 20 - x$.

$12x + 7(20 - x) = 180 \;\Rightarrow\; 5x = 40 \;\Rightarrow\; x = 8$

The blend uses $8$ pounds of the $12$-dollar coffee.

Downstream $2(b + c) = 12$ gives $b + c = 6$; upstream $3(b - c) = 12$ gives $b - c = 4$. Subtract to eliminate $b$.

$(b + c) - (b - c) = 6 - 4 \;\Rightarrow\; 2c = 2 \;\Rightarrow\; c = 1$

The current runs at $1$ mph (and the canoe at $5$ mph).

Let $L$ and $W$ be length and width: $2L + 2W = 54$, so $L + W = 27$, and $L = 2W + 3$. Substitute.

$(2W + 3) + W = 27 \;\Rightarrow\; 3W = 24 \;\Rightarrow\; W = 8$

Then $L = 2(8) + 3 = 19$, so the length is $19$ cm.

With the wind the ground speed is $2100 / 3 = 700$, so $p + w = 700$; against it $2100 / 3.5 = 600$, so $p - w = 600$. Subtract to eliminate $p$.

$(p + w) - (p - w) = 700 - 600 \;\Rightarrow\; 2w = 100 \;\Rightarrow\; w = 50$

The wind speed is $50$ mph (and the jet is $650$ mph in still air).

Let $x$ be the liters of $15\%$ solution and $y$ the liters of $40\%$: $x + y = 10$ and $0.15x + 0.40y = 0.25(10) = 2.5$. Substitute $x = 10 - y$.

$0.15(10 - y) + 0.40y = 2.5 \;\Rightarrow\; 0.25y = 1 \;\Rightarrow\; y = 4$

Then $x = 10 - 4 = 6$, so $6$ liters of the $15\%$ solution (and $4$ of the $40\%$).

Work in cents. Let $n$ and $q$ be the counts: $n + q = 22$ and $5n + 25q = 390$. Multiply the count equation by $5$ and subtract.

$(5n + 25q) - (5n + 5q) = 390 - 110 \;\Rightarrow\; 20q = 280 \;\Rightarrow\; q = 14$

There are $14$ quarters (and $8$ nickels).

Let $M$ and $T$ be their present ages: $M = 3T$ and $M + 4 = 2(T + 4)$. Substitute $3T$ for $M$.

$3T + 4 = 2T + 8 \;\Rightarrow\; T = 4$

Tom is $4$ now (and Maria is $12$).

Let $b$ be the still-water speed and $c$ the current, both unknown. Downstream $2(b + c) = 16$ gives $b + c = 8$; upstream $4(b - c) = 16$ gives $b - c = 4$. Add to eliminate $c$.

$2b = 12 \;\Rightarrow\; b = 6$

The still-water speed is $6$ mph (and the current is $2$ mph).

Work in cents. Let $q$ be quarters and $d$ dimes: $d = 2q$ and $25q + 10d = 450$. Substitute $2q$ for $d$.

$25q + 10(2q) = 450 \;\Rightarrow\; 45q = 450 \;\Rightarrow\; q = 10$

There are $10$ quarters (and $20$ dimes).

One equation adds the volumes and one adds the juice content, matching each concentration to its own variable.

$x + y = 14, \qquad 0.25x + 0.60y = 0.40(14)$

Using whole percents as $25x + 60y$ mixes up the units, and pairing $0.60$ with $x$ swaps the two concentrations.