Equations with Two Variables: Additional Practice (Set 1)

Substitute $y = 3$ into $3x + 4y = 24$ and solve.

$3x + 12 = 24, \qquad 3x = 12, \qquad x = 4$

Substitute $x = k$ and $y = 2k$ into $x + 2y = 15$ and combine.

$k + 2(2k) = 15, \qquad 5k = 15, \qquad k = 3$

Both points must satisfy the equation, so test each in the candidate.

$2(1) - 1 = 1 \quad\text{and}\quad 2(4) - 7 = 1$

Only $2x - y = 1$ holds for both points.

Substitute $x = -3$ into $2x + y = -1$ and solve.

$2(-3) + k = -1, \qquad -6 + k = -1, \qquad k = 5$

Every pair listed adds to $20$, so test the second condition, that the first number is three times the second, on each candidate.

$15 = 3 \times 5$

Only $(15, 5)$ passes, since $12 \neq 3(8)$, $16 \neq 3(4)$, and $10 \neq 3(10)$.

Move the $x$ term across to reach $ax + by = c$ with a positive leading coefficient.

$y = -2x + 6 \;\Rightarrow\; 2x + y = 6$

Set $y = x$ in $6x + 4y = 24$ and combine like terms.

$6x + 4x = 24, \qquad 10x = 24, \qquad x = 2.4$

So the pair with equal coordinates is $(2.4, 2.4)$.

Substitute the pair $(2, 5)$ into $ax + y = 11$ and solve for $a$.

$2a + 5 = 11, \qquad 2a = 6, \qquad a = 3$

The pair must add to $6$ with the first coordinate larger than the second.

$4 + 2 = 6 \quad\text{with}\quad 4 > 2$

So $(4, 2)$ fits. The others either reverse the inequality or make the coordinates equal.

A linear equation keeps each variable to the first power and never multiplies the two variables together.

$xy = 12$

Here $x$ and $y$ are multiplied, so it is not linear. The other three are linear (a coefficient of $\tfrac{1}{2}$ is still linear).

Substitute $x = 6$ into $2x + 3y = 6$ and solve, expecting a negative value.

$12 + 3y = 6, \qquad 3y = -6, \qquad y = -2$

Substitute $x = 4$ into $y = 2x + 3$.

$y = 2(4) + 3 = 8 + 3 = 11$

So a 4-mile ride costs $11$ dollars.