$6 = \sqrt{2 z - 3} - 3 ?$ $6 = sqrt( 2z - 3) - 3?$ solve the equation

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Answer 1 · 2017-04-11T05:39:55

$z = 42$ $z = 42$

Assuming your question is $6 = \sqrt{2 z - 3} - 3$ $6 = sqrt(2 z -3) -3$

add $3$ $3$ to both sides
$6 + 3 = \sqrt{2 z - 3} - 3 + 3$ $6 + 3= sqrt(2 z -3) -3 + 3$
$9 = \sqrt{2 z - 3}$ $9= sqrt(2 z -3)$

square to both sides
$9^{2} = {(\sqrt{2 z - 3})}^{2}$ $9^2= (sqrt(2 z -3))^2$
$81 = 2 z - 3$ $81 = 2 z - 3$

add $3$ $3$ to both sides
$81 + 3 = 2 z - 3 + 3$ $81 + 3= 2 z - 3 + 3$
$84 = 2 z$ $84 = 2 z$

divide $2$ $2$ to both sides
$\frac{84}{2} = \frac{2 z}{2}$ $84/2 = (2 z) /2$

$42 = z$ $42 = z$

6=√2z−3−3?6 = sqrt( 2z - 3) - 3? solve the equation