Question

solve each of the following equations and verify the answer in each case.
$16(3x - 5) - 10(4x - 8) = 40$
$3(x + 6) + 2(x + 3) = 64$
$3(x + 6) + 2(1 - 6x) = 1$
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Answer 1

Step-by-step explanation:

the cases are

1. b²-4ac=0
2. b²-4ac<0
3. b²-4ac>0

Answer 2

Solution

$\sf\bold{16(3x - 5) - 10(4x - 8) = 40}$

$\sf\longmapsto16(3x - 5) - 10(4x - 8) = 40$

$\sf\longmapsto48x - 80 - 40x + 80 = 40$

$\sf\longmapsto48x - 40 = 40$

$\sf\longmapsto8x = 40$

$\sf\bold{\longmapsto \: x = 5}$

Verify

$\sf\longmapsto16(15 - 5) - 10(20 - 8)$

$\sf\longmapsto16 \times 10 - 10(12)$

$\longmapsto160 - 120 = 40$

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$\sf\longmapsto3(x + 6) + 2(x + 3) = 64$

$\sf\longmapsto3x + 18 + 2x + 6 = 64$

$\sf\longmapsto5x + 24 = 64$

$\sf\longmapsto5x = 64 - 24 = 40$

$\sf\bold{\longmapsto \: x = 8}$

Verify

$\sf\longmapsto3(14) + 2(11)$

$\sf\longmapsto42 + 22 = 64$

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$\sf\longmapsto3(x + 6) +2(1 - 6x) = 1$

$\sf\longmapsto3x + 18 + 2 - 12x = 1$

$\sf\longmapsto - 9x + 20 = 1$

$\sf\longmapsto - 9x = - 19$

$\sf\longmapsto \: x = \dfrac{19}{9}$

Verify

$\sf\longmapsto3\bigg( \dfrac{19}{9} + 6\bigg) + 2\bigg(1 - 6 \times \dfrac{19}{9} \bigg)$

$\sf\longmapsto3\bigg( \dfrac{19 + 54}{9} \bigg) + 2\bigg( \dfrac{9 - 114}{9} \bigg)$

$\sf\longmapsto3\bigg( \dfrac{73}{9}\bigg ) + 2\bigg( - \dfrac{105}{9} \bigg)$

$\sf\longmapsto \dfrac{219}{9} - \dfrac{210}{9} = \dfrac{9}{9} = 1$