Question

this is an example of an sat/tsi questions. no graphing allowed.
if $sqrt{x+z} = sqrt{3x+10}$, find the solution if the value of $z = 8$.
$circ$ $x = 3$
$circ$ $x = -1$
$circ$ $x = -2$
$circ$ $x = 5$

Explanation:

Step1: Substitute z=8 into equation

$\sqrt{x+8} = \sqrt{3x+10}$

Step2: Square both sides to eliminate roots

$(\sqrt{x+8})^2 = (\sqrt{3x+10})^2$
$x+8 = 3x+10$

Step3: Rearrange to solve for x

$8-10 = 3x-x$
$-2 = 2x$
$x = \frac{-2}{2} = -1$

Step4: Verify the solution

Left side: $\sqrt{-1+8}=\sqrt{7}$
Right side: $\sqrt{3(-1)+10}=\sqrt{7}$
Both sides are equal, so x=-1 is valid.

Answer:

B. x = -1