Equations with Two Variables: Practice

Double the first coordinate and subtract the second, looking for $5$.

$2(4) - 3 = 8 - 3 = 5$

So $(4, 3)$ is a solution. The others give $2$, $3$, and $6$.

Subtract $2x$ from both sides to leave $y$ alone.

$y = 7 - 2x$

The $2x$ crosses the equals sign as $-2x$.

Substitute $y = 3$ into $2x + 3y = 19$ and solve for $k$.

$2k + 3(3) = 19, \qquad 2k + 9 = 19, \qquad 2k = 10, \qquad k = 5$

Substitute $x = 3$ into $4x + 2y = 20$ and solve.

$4(3) + 2y = 20, \qquad 12 + 2y = 20, \qquad 2y = 8, \qquad y = 4$

Add the first coordinate to twice the second and look for $0$.

$2 + 2(-1) = 2 - 2 = 0$

So $(2, -1)$ is a solution. The others give $5$, $4$, and $3$.

Move the $x$ term across to reach the form $ax + by = c$. Subtract $2x$ from both sides, then tidy the signs.

$y - 2x = 3 \;\Rightarrow\; 2x - y = -3$

Multiplying $-2x + y = 3$ through by $-1$ gives $2x - y = -3$.

Triple the first coordinate, subtract twice the second, and look for $6$.

$3(4) - 2(3) = 12 - 6 = 6$

So $(4, 3)$ is a solution. The others give $0$, $-6$, and $4$.

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

$y = 3(2) - 4 = 6 - 4 = 2$

Two pens at $x$ dollars each cost $2x$, and three notebooks at $y$ dollars each cost $3y$. Together they come to $13$ dollars.

$2x + 3y = 13$

Solving for $y$ gives $y = 9 - 3x$. Substitute $x = -1$, watching the double negative.

$y = 9 - 3(-1) = 9 + 3 = 12$

Solving for $y$ gives one matching value for every choice of $x$.

$y = \frac{17 - 2x}{5}$

With endlessly many choices for $x$, the equation has infinitely many solutions.

A point lies on the line when its coordinates add to $8$, so the exception is the pair that does not.

$5 + 4 = 9 \neq 8$

So $(5, 4)$ is not on the line. The other three each add to $8$.