Question

Fine the x and y intercepts of the following equation. Fill in blank one and blank two.

Answer 1

EXPLANATION:

We are given the following linear equation;

$2x+3y=12$

We would begin by re-writing the equation in the slope-intercept form as shown below;

$y=mx+b$

We now have;

$\begin{gathered} 2x+3y=12 \\ Make\text{ y the subject of the equation;} \end{gathered}$$3y=12-2x$$\begin{gathered} Divide\text{ all through by 3;} \\ \frac{3y}{3}=\frac{12}{3}-\frac{2x}{3} \end{gathered}$$y=4-\frac{2}{3}x$

We can re-arrange this in the format shown earlier;

$y=-\frac{2}{3}x+4$

The x-intercept is derived at the point where y = 0. Same for the y-intercept. Its derived at the point where x = 0.

Hence, for the equation;

$\begin{gathered} y=-\frac{2}{3}x+4 \\ When\text{ }x=0 \\ y=-\frac{2}{3}(0)+4 \\ y=4 \end{gathered}$$\begin{gathered} When\text{ }y=0 \\ 0=-\frac{2}{3}x+4 \\ \frac{2}{3}x=4 \end{gathered}$

Cross multiply and we'll have;

$\begin{gathered} 2x=3\times4 \\ 2x=12 \\ \end{gathered}$

Divide both sides by 2;

$x=6$

Therefore, the x and y intercepts are;

ANSWER:

$\begin{gathered} x-intercept=6 \\ y-intercept=4 \end{gathered}$