Question

Solve each of the following equations and also check your results in each case:

$\left(xxiv\right)$ $\begin{array}{c}(3x-8)(3x+2)-(4x-11)\\ (2x+1)=(x-3)(x+7)\end{array}$

Answer 1

Given: $(3x–8)(3x+2)–(4x–11)(2x+1)=$
$(x–3)(x+7)$

$9x^2+6x–24x–16–8x^2–4x+22x+11$
$=x^2+7x–3x–21$

$9x^2+6x–24x–16–8x^2–4x+22x$
$+11–x^2–7x+3x+21=0$

$9x^2–8x^2–x^2+6x–24x–4x+22x–7x$
$+3x–16+21+11=0$

$-4x+16=0$

$-4x=-16$

$x=4$

To check: $3x–8\left)\right(3x+2)–(4x–11\left)\right(2x+1)$
$=(x–3)(x+7)$ for $x=4$

L.H.S $=(3x–8)(3x+2)–(4x–11)(2x+1)$

$=\left[3\right(4)–8]\left[3\right(4)+2]–\left[4\right(4)–11]\left[\right(2\left(4\right)+1]$

$=(12-8)(12+2)–(16-11)(8+1)$

$=4\left(14\right)–5\left(9\right)$

$=56–45=11$

R.H.S $=(x–3)(x+7)$

$=(4–3)(4+7)$

$=1\left(11\right)$

$=11$

$\therefore \text{L.H.S = R.H.S}$

Hence, the given equation is verified for $x=4.$