# Given h(x)=4x−5, how do you solve for x when h(x)=−1?

Jan 6, 2017

$x = 1$

#### Explanation:

$h \left(x\right) = 4 x - 5 , h \left(x\right) = - 1$

replace $h \left(x\right)$ with $- 1$ in the function

$\implies - 1 = 4 x - 5$

we now solve for $x$

add $5$ to both sides

$- 1 + 5 = 5 x \cancel{- 5 + 5}$

$4 = 4 x$

$\implies x = 1$