Word Problems with Systems: Practice

Let $x$ be the larger and $y$ the smaller: $x + y = 50$ and $x - y = 12$. Add the equations to eliminate $y$.

$2x = 62 \;\Rightarrow\; x = 31$

The larger number is $31$ (and the smaller is $19$).

Let $f$ and $b$ be the floor and balcony counts: $f + b = 300$ and $18f + 12b = 4200$. Multiply the count equation by $12$ and subtract to remove $b$.

$(18f + 12b) - (12f + 12b) = 4200 - 3600 \;\Rightarrow\; 6f = 600 \;\Rightarrow\; f = 100$

So $100$ floor tickets (and $200$ balcony) were sold.

Let $q$ and $d$ be the counts: $q + d = 40$ and $25q + 10d = 700$. Multiply the count equation by $10$ and subtract.

$(25q + 10d) - (10q + 10d) = 700 - 400 \;\Rightarrow\; 15q = 300 \;\Rightarrow\; q = 20$

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

Let $x$ be the liters of $10\%$ solution and $y$ the liters of $30\%$: $x + y = 20$ and $0.10x + 0.30y = 0.25(20) = 5$. Substitute $x = 20 - y$.

$0.10(20 - y) + 0.30y = 5 \;\Rightarrow\; 0.20y = 3 \;\Rightarrow\; y = 15$

Then $x = 20 - 15 = 5$, so $5$ liters of the $10\%$ solution.

Downstream $2(b + c) = 40$ gives $b + c = 20$; upstream $4(b - c) = 40$ gives $b - c = 10$. Add to eliminate $c$.

$2b = 30 \;\Rightarrow\; b = 15$

The still-water speed is $15$ mph (and the current is $5$ mph).

Let $p$ be pencils and $q$ pens: $p + q = 12$ and $2p + 5q = 39$. Substitute $p = 12 - q$.

$2(12 - q) + 5q = 39 \;\Rightarrow\; 24 + 3q = 39 \;\Rightarrow\; q = 5$

You bought $5$ pens (and $7$ pencils).

Let $c$ be pounds of cashews and $r$ pounds of raisins: $c + r = 10$ and $9c + 3r = 6(10) = 60$. Substitute $r = 10 - c$.

$9c + 3(10 - c) = 60 \;\Rightarrow\; 6c = 30 \;\Rightarrow\; c = 5$

The mix uses $5$ pounds of cashews (and $5$ of raisins).

Each quarter is $25$ cents and each dime $10$ cents.

$25(7) + 10(13) = 175 + 130 = 305$

The coins are worth $305$ cents.

First find break-even by setting cost equal to revenue: $5x = 2x + 300$, so $3x = 300$ and $x = 100$. Then the revenue is $5x$.

$y = 5(100) = 500$

The revenue at break-even is $500$ dollars.

Multiply the count equation by $5$ and subtract to eliminate $s$.

$(8a + 5s) - (5a + 5s) = 1700 - 1250 \;\Rightarrow\; 3a = 450 \;\Rightarrow\; a = 150$

So $150$ adult tickets (and $100$ student) were sold.

With the wind $3(p + w) = 1500$ gives $p + w = 500$; against it $5(p - w) = 1500$ gives $p - w = 300$. Add to eliminate $w$.

$2p = 800 \;\Rightarrow\; p = 400$

The still-air speed is $400$ mph (and the wind is $100$ mph).

Let $x$ be the $50$-cent stamps and $y$ the $20$-cent stamps: $x + y = 30$ and $50x + 20y = 900$. Multiply the count equation by $20$ and subtract.

$(50x + 20y) - (20x + 20y) = 900 - 600 \;\Rightarrow\; 30x = 300 \;\Rightarrow\; x = 10$

There are $10$ stamps of $50$ cents (and $20$ of $20$ cents).