Graphing Linear Equations: Additional Practice (Set 2)

A graph passes through the origin when $(0, 0)$ satisfies its equation.

$y = 3x: \quad 0 = 3(0) = 0$

So $y = 3x$ passes through the origin. The others give $2$, $4$, and $-1$ at the origin, not $0$.

The three points $(0, 0)$, $(1, 3)$, and $(2, 6)$ line up evenly along one straight line, and $(1, 3)$ sits exactly halfway between $(0, 0)$ and $(2, 6)$: its $x$-coordinate is halfway from $0$ to $2$, and its $y$-coordinate is halfway from $0$ to $6$.

$(1,\, 3)$

So $(1, 3)$ lies on the line, while the other choices do not.

The y-intercept is where the line meets the vertical axis, so read where it crosses the y-axis.

$\text{The crossing sits at } x = 0, \; y = 3$

So the y-intercept is $(0, 3)$. The point $(4, 0)$ is the x-intercept.

Substitute $x = 3$ into $x + y = 7$ and solve for $y$.

$3 + y = 7, \qquad y = 4$

So the point on the line is $(3, 4)$.

Substitute $x = k$ and $y = k$ into $x + y = 10$.

$k + k = 10, \qquad 2k = 10, \qquad k = 5$

So the point is $(5, 5)$.

The point must satisfy both rules at once: $x = 5$ and $y = 2$.

$(x, y) = (5, 2)$

So the two lines cross at $(5, 2)$.

Set $y = 0$ for the x-intercept: $2x = 6$ gives $x = 3$, the point $(3, 0)$. Set $x = 0$ for the y-intercept: $y = 6$, the point $(0, 6)$.

$(3, 0) \text{ and } (0, 6)$

So the intercepts are $(3, 0)$ and $(0, 6)$.

At the x-intercept $y = 0$, so substitute and solve for $x$.

$3x - 0 = 9, \qquad 3x = 9, \qquad x = 3$

So $a = 3$ and the x-intercept is $(3, 0)$.

From $x = 1$ to $x = 2$ the value of $y$ rises from $4$ to $7$, a step of $3$ for each step of $1$ in $x$. The same step applies again from $x = 2$ to $x = 3$.

$y = 7 + 3 = 10$

So the row at $x = 3$ is $(3, 10)$.

Test $(2, 6)$ in each equation. For $x = 3$:

$x = 2 \neq 3$

So $x = 3$ does not pass through $(2, 6)$. The other three all hold: $3(2) = 6$, $2 + 6 = 8$, and $y = 6$.

Test $(0, 0)$ in each equation; it passes through the origin only if both sides match.

$y = x + 1: \quad 0 = 0 + 1 = 1 \; \text{(false)}$

So $y = x + 1$ does not pass through the origin. The other three all give $0 = 0$ there.

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

$y = -(0) + 4 = 4$

So the graph crosses the y-axis at $(0, 4)$.