Graphing Linear Equations: Additional Practice (Set 1)

Set $y = 0$ and solve for $x$.

$7x + 4(0) = 28, \qquad 7x = 28, \qquad x = 4$

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

The line is straight up and down, and every marked point sits in the same column, three units left of the y-axis.

$\text{every point has } x = -3$

So the equation is $x = -3$. A vertical line fixes $x$ and leaves $y$ free.

Substitute $x = -1$ into $y = 4x + 6$, keeping the sign.

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

So $k = 2$ and the point is $(-1, 2)$.

Substitute each pair into $5x - 2y = 10$. For $(0, -5)$:

$5(0) - 2(-5) = 0 + 10 = 10$

So $(0, -5)$ is on the line. The others give $-10$, $0$, and $21$.

Set $x = 0$ and solve for $y$.

$6(0) + 5y = 20, \qquad 5y = 20, \qquad y = 4$

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

The vertical line $x = -1$ meets the x-axis where $y = 0$, keeping $x = -1$.

$(x, y) = (-1, 0)$

So it crosses the x-axis at $(-1, 0)$.

The x-intercept has $y = 0$, so set $y = 0$ and solve.

$0 = 3x - 6, \qquad 3x = 6, \qquad x = 2$

So the graph crosses the x-axis at $(2, 0)$.

A line parallel to the x-axis is horizontal, coming from an equation that fixes $y$ at a constant.

$y = 7 \;\Rightarrow\; \text{a horizontal line}$

Here $x = 2$ and $x = -5$ are vertical, and $y = x$ is slanted.

Set $x = 0$ and solve for $y$.

$2(0) - 7y = 14, \qquad -7y = 14, \qquad y = -2$

So the y-intercept is $(0, -2)$.

Set $y = 0$ for the x-intercept: $4x = 12$ gives $x = 3$, the point $(3, 0)$. Set $x = 0$ for the y-intercept: $3y = 12$ gives $y = 4$, the point $(0, 4)$.

$(3, 0) \text{ and } (0, 4)$

So the intercepts are $(3, 0)$ and $(0, 4)$.

Both points have the same y-coordinate, $2$, so every point along the segment stays at height $2$.

$y = 2 \text{ for both points}$

Equal heights give a horizontal line, $y = 2$.

Set $x = 0$ and solve for $y$.

$5(0) + 2y = 20, \qquad 2y = 20, \qquad y = 10$

So the y-intercept is $(0, 10)$.