Question
Solve the following equation and also check your result in case: $(3\text{x}-8)(3\text{x+2})-(4\text{x}-11)(2\text{x}+1)=(\text{x}-3)(\text{x}+7)$
✓
Answer
$(3\text{x}-8)(3\text{x+2})-(4\text{x}-11)(2\text{x}+1)=(\text{x}-3)(\text{x}+7)$
$9\text{x}^2+6\text{x}-24\text{x}-16-8\text{x}^2-4\text{x}+22\text{x}+11$
$=\text{x}^2+7\text{x}-3\text{x}-21$
$\text{x}^2-5=\text{x}^2+4\text{x}-21$
$4\text{x}=-5+21$
$\text{x}=\frac{16}{4}=4$
Thus, $\text{x}=4$ is the solution of the given equation.
Check:Substituting $\text{x}=4$ in the given equation, we get:
$\text{L.H.S.}=(3\times4-8)(3\times4+2)-(4\times4-11)(2\times4+1)$
$=4\times14-5\times9=11$
$\text{R.H.S.}=(4-3)(4+7)=11$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=4$
$9\text{x}^2+6\text{x}-24\text{x}-16-8\text{x}^2-4\text{x}+22\text{x}+11$
$=\text{x}^2+7\text{x}-3\text{x}-21$
$\text{x}^2-5=\text{x}^2+4\text{x}-21$
$4\text{x}=-5+21$
$\text{x}=\frac{16}{4}=4$
Thus, $\text{x}=4$ is the solution of the given equation.
Check:Substituting $\text{x}=4$ in the given equation, we get:
$\text{L.H.S.}=(3\times4-8)(3\times4+2)-(4\times4-11)(2\times4+1)$
$=4\times14-5\times9=11$
$\text{R.H.S.}=(4-3)(4+7)=11$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=4$
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