How do you solve $12x+132=12x-100$ by graphing?

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Answer 1 · 2017-01-23T12:05:41

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There is no common point so no solution to this system of equations as presented.

$color(blue)("Important point to note before we start")$

Given: $12x+132=12x-100$

Subtract $12x$ from both sides

$132=-100 color(red)(larr" This is untrue so no solution")$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Answering the question as if there is a system of equations")$

In circumstances such as this you would set both as equal to $y$ so that you have the format of:

$12x+132=y_1=12x-100 color(red)(larr" This proves to be false")$

However, take a closer look at it. Notice that they both have

$12x+" some constant"$

Now consider the standardised form:

$y=mx+c$ where $m$ is the gradient.

$y_2=12x+132" "y_3=12x-100$
$y=mx+c" "y=mx+c$

The gradient of $m$ is the same in both equations. So they both have the same gradient (slope)

$color(red)("They are parallel")$

As they have different starting points on the y-axis ( y-intercept ) they do not share a common point anywhere.

$color(red)("Thus there is no solution")$

Tony B

How do you solve 12x+132=12x-10012x+132=12x-100 by graphing?