#### Answer

$x=-4$ and $x=\dfrac{1}{2}$ satisfy the given conditions

#### Work Step by Step

$y_{1}=2x+3$ $,$ $y_{2}=x+2$ and $y_{1}y_{2}=10$
If $y_{1}y_{2}$ is equal to $10$, then $(2x+3)(x+2)$ is also equal to $10$
$(2x+3)(x+2)=10$
Evaluate the product on the left side:
$2x^{2}+4x+3x+6=10$
Take the $10$ to the left side and simplify:
$2x^{2}+4x+3x+6-10=0$
$2x^{2}+7x-4=0$
Solve by factoring:
$(x+4)(2x-1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+4=0$
$x=-4$
$2x-1=0$
$2x=1$
$x=\dfrac{1}{2}$
$x=-4$ and $x=\dfrac{1}{2}$ satisfy the given conditions