Graphing Linear Equations: Practice

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

$3x + 5(0) = 15, \qquad 3x = 15, \qquad x = 5$

So the x-intercept is $(5, 0)$. Setting $x = 0$ instead gives the y-intercept $(0, 3)$, a tempting wrong choice.

Test the two marked points in each equation and keep the one both satisfy. For $y = 2x - 1$:

$x = 0:\; y = -1, \qquad x = 2:\; y = 2(2) - 1 = 3$

Both marked points fit, so the line is $y = 2x - 1$. Each other equation misses at least one point.

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

$4(0) + 3y = 9, \qquad 3y = 9, \qquad y = 3$

So the y-intercept is $(0, 3)$.

Because the point is on the line, substitute $x = 6$ and solve for $y$.

$2(6) - 3y = 3, \qquad 12 - 3y = 3, \qquad -3y = -9, \qquad y = 3$

So $k = 3$ and the point is $(6, 3)$.

A point is on the line when its coordinates satisfy $y = 2x - 4$. Check $(2, -1)$:

$2(2) - 4 = 0 \neq -1$

So $(2, -1)$ is not on the line. The other three each satisfy the equation.

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

$0 = 5 - 2x, \qquad 2x = 5, \qquad x = \tfrac{5}{2}$

So the graph crosses the x-axis at $\left(\tfrac{5}{2}, 0\right)$. An intercept need not land on a whole-number mark.

A vertical line fixes $x$, and it passes through $(3, -2)$ only if that fixed value is $3$.

$x = 3 \;\Rightarrow\; \text{vertical line through every point with } x = 3$

So the answer is $x = 3$. The equation $y = -2$ is the horizontal line through the same point.

First simplify $2y = 8$ by dividing both sides by $2$.

$y = 4$

So the graph is the horizontal line $y = 4$, every point at height $4$.

The line $x = 0$ is the y-axis: every point on it has x-coordinate $0$.

$(0, -3) \;\Rightarrow\; x = 0$

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

A point is on the graph when it satisfies $y = 3x + 2$. Check $(2, 8)$:

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

So $(2, 8)$ is on the line. The others give $5$, $2$, and $11$.

The horizontal line through $(2, -5)$ is $y = -5$: every point on it has y-coordinate $-5$, whatever its x-coordinate.

$(7, -5) \;\Rightarrow\; y = -5$

So $(7, -5)$ is on it. The others do not have $y = -5$.

The crossing point must satisfy both rules: $x = 2$ and $y = -3$ at once.

$(x, y) = (2, -3)$

So the horizontal line $y = -3$ and the vertical line $x = 2$ cross at $(2, -3)$.