How do you solve and graph abs(-4y-3)<13?

Jan 5, 2018

It's about knowing the meaning of 'absolute value' or in other words the distance from zero in a axis.

Explanation:

So let's first look a simple example. How do we figure out the value of x for which the absolute value of x+3 is greater than 6.

Making this in mathematical format, the question becomes:

$\left\mid x + 3 \right\mid > 6$

If $\left\mid x + 3 \right\mid = 6$, what would be the value of x?

It would be either $x + 3 = 6$ or $x + 3 = - 6$...since if $\left\mid x \right\mid = 6$ it actually means the distance from x to zero is 6 units. On a axis x could be either -6 or 6, so x could attain a distance of 6.

Know let's go back to your question i.e $\left\mid - 4 y - 3 \right\mid < 13$. As we seen before here instead of just a single variable like x or y we have an expression. So instead of the variable being the distance from zero here the expression's value is the distance from zero.

Know the expression $- 4 y - 13$ would have a distance of 13 from zero. So again either $- 4 y - 13$ or $- \left(4 y - 13\right)$, so the expression $- 4 y - 13$ could attain a distance of 13.

Expressing mathematically:

$- 4 y - 13 < 13$ or $- 4 y + 13 < 13$
$- 4 y < 26$ or $- 4 y < 0$
$y > - \frac{26}{4}$ or $y > 0$ which would be the solution.