Equations with Two Variables: Practice

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

$y = 3 + 2 = 5$

So $y = 5$.

Substitute each pair into $2x + y = 6$, doubling the first coordinate and adding the second.

$2(1) + 4 = 2 + 4 = 6$

So $(1, 4)$ is a solution. The others give $7$, $5$, and $9$.

Double the first coordinate, add the second, and look for $8$.

$2(3) + 2 = 6 + 2 = 8$

So $(3, 2)$ is a solution. The others give $6$, $10$, and $4$.

Substitute $x = 5$ into $x - y = 2$ and solve for $y$.

$5 - y = 2, \qquad y = 3$

A linear equation in two variables keeps each variable to the first power: the variables are not multiplied together, squared, or placed in a denominator.

$2x + y = 5$

Only this one qualifies. In $xy = 5$ the variables are multiplied, $x^2 + y = 5$ squares $x$, and $\dfrac{1}{x} + y = 5$ puts $x$ in a denominator.

For any value you choose for $x$, the equation gives exactly one matching $y$, namely $y = 4 - x$.

$y = 4 - x$

Since there are endlessly many choices for $x$, there are infinitely many solutions.

A pair $(x, y)$ solves $y = x - 1$ when the second coordinate is one less than the first.

$4 = 5 - 1$

So $(5, 4)$ works. The others fail: $5 \neq 4 - 1$, $3 \neq 3 - 1$, and $3 \neq 2 - 1$.

Add the coordinates of each pair; a solution totals $6$, so the exception is the pair that does not.

$4 + 3 = 7 \neq 6$

So $(4, 3)$ is not a solution. The other three each add to $6$.

Substitute $x = 3$ into $x + y = 10$ and solve for the missing coordinate.

$3 + k = 10, \qquad k = 7$

Substitute $y = 2$ into $x + y = 9$ and solve for $k$.

$k + 2 = 9, \qquad k = 7$

Substitute $x = 0$ into $2x + y = 5$. The term $2x$ becomes $0$ but is still part of the equation.

$2(0) + y = 5, \qquad y = 5$

Subtract the second coordinate from the first and look for $1$.

$4 - 3 = 1$

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