Graphing Linear Equations: Practice

A pair is a solution when its y-coordinate equals its x-coordinate plus $3$.

$5 = 2 + 3$

So $(2, 5)$ works. The others fail, since $4 + 3 = 7$, $5 + 3 = 8$, and $1 + 3 = 4$.

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

$y = 4 + 3 = 7$

So $y = 7$, giving the solution point $(4, 7)$.

A point is on the graph when its y-coordinate is twice its x-coordinate.

$6 = 2 \times 3$

So $(3, 6)$ lies on the line. The others do not: $2(6) = 12$, $2(2) = 4$, and $2(5) = 10$.

The equation $y = 4$ fixes the height at $4$ and puts no condition on $x$, so $x$ can be anything while $y$ stays $4$.

$(x,\, 4) \text{ for every value of } x$

All these points sit at height $4$, forming a horizontal line.

Substitute $x = 3$ into $y = 2x - 1$.

$y = 2(3) - 1 = 6 - 1 = 5$

So $y = 5$, giving the point $(3, 5)$.

Every point on the line $x = 5$ has x-coordinate $5$, whatever its y-coordinate.

$(5,\, 2) \;\Rightarrow\; x = 5$

So $(5, 2)$ is on the line. The others have x-coordinates $2$, $0$, and $3$.

A table pair must satisfy $y = x - 2$. Take $x = 5$:

$y = 5 - 2 = 3$

So $(5, 3)$ belongs in the table.

The x-intercept is where the graph crosses the x-axis, where $y = 0$. Set $y = 0$ and solve for $x$.

$0 = x - 4, \qquad x = 4$

So the x-intercept is $(4, 0)$.

A point is on the graph when it satisfies $y = 2x + 1$. Check $(2, 5)$:

$2(2) + 1 = 5$

So $(2, 5)$ is on the line; the others fail the equation.

Set $x = 0$ to find where the graph meets the y-axis.

$y = 3(0) = 0$

So the y-intercept is $(0, 0)$: this line passes through the origin.

Double the first coordinate, add the second, and look for $7$.

$2(2) + 3 = 4 + 3 = 7$

So $(2, 3)$ is a solution. The others give $5$, $9$, and $8$.

On the line $y = x$ the two coordinates are equal.

$7 = 7$

So $(7, 7)$ is on the line; the others have unequal coordinates.