How do you solve (11x - 5) ^ { 2} - ( 10x - 1) ^ { 2} - ( 3x - 20) ( 7x + 10) = 124?

Jun 12, 2018

color(crimson)(x = 40/53

Explanation:

Follow the PEMDAS Rule to solve.

${\left(11 x - 5\right)}^{2} - {\left(10 x - 1\right)}^{2} - \left(3 x - 2\right) \left(7 x + 10\right) = 124$

121x^2 - 110x + 25 - 100x^2 + 20x - 1 - 21x^2 - 30x + 14x + 20 = 124, color(purple)(" Removing Braces"

121x^2 - 100x^2 - 21x^2 - 110x + 20x - 30x + 14x + 25 - 1 + 20 = 124, color(purple)("Bringing like terms together"

0 x^2 - 106x + 44 = 124, color(purple)("Adding / Subtracting"

$106 x = - 124 + 44$

$106 x = - 80$

color(crimson)(x = 80/106 " or " x = 40/53, color(purple)("Simplifying"

Jun 12, 2018

$x = - 5$

Explanation:

We have

${\left(11 x - 5\right)}^{2} - {\left(10 x - 1\right)}^{2}$
$= \left(11 x - 5 - 10 x + 1\right) \left(11 x - 5 + 10 x - 1\right)$
$= \left(x - 4\right) \left(21 x - 6\right) = 21 {x}^{2} - 90 x + 24$

and

$\left(3 x - 20\right) \left(7 x + 10\right) = 21 {x}^{2} - 110 x - 200$

Thus, the equation is

$\left(21 {x}^{2} - 90 x + 24\right) - \left(21 {x}^{2} - 110 x - 200\right) = 124 \implies$

$20 x + 100 = 0 \implies 20 \left(x + 5\right) = 0 \implies$

$x = - 5$