Question

When I try to find x and y intercepts, why do I get double the number it's supposed to be? Here's the problem I'm working with.
3x + 4y = 24
This was my solution when I let 0 represent x when finding the y intercept, and when finding the x intercept i let 0 represent y.
3*8+4*6=48
Is this the correct answer? If not what did I do wrong?

Answer 1

You're finding the y intercept so,

3(0) + 4y = 2, you know that 3(0) is 3*0 which is 0 because anything times 0 will always be 0.

So now solve for y, 4y = 24, divide both sides by 4 and you get, \(\ \sf \dfrac{\cancel{4}y}{\cancel{4}} = \dfrac{24}{4} \longrightarrow y = 6 \) So (0,6) :P

Answer 2

I've got that part down. But when I verified the solution by substituting the values into the original equations, I got this 3*8+4*6=48
Is this correct? @tHe_FiZiCx99

Answer 3

The point is that you want to find the intercepts,

They're suppose to be done separately,

x = 8 and y = 6

When you plug those in you will get 24.

Answer 4

Btw, you don't let 0 represent y, you plug in 0 FOR y.