Question #09738

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Question #09738

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Answer 1 · 2017-09-11T22:18:09

$13 a - 2$ $color(blue)(13a - 2 )$

Explanation:

$16 (a - 2) + 3 (10 - a)$ $16( a - 2 ) + 3( 10 - a )$

Multiply brackets by the value outside of them to remove them.

$16 a - 32 + 30 - 3 a$ $16a - 32 + 30 -3a$

Collect like terms:

$16 a - 3 a - 32 + 30 = 13 a - 2$ $16a -3a -32 + 30 = color(blue)(13a -2)$

Answer 2 · 2017-09-11T22:18:37

$= 13 a - 2$ $=13a-2$

Explanation:

We start using the expansion method where:
$a (b + c) = a b + a c$ $a(b+c)=ab+ac$

$\to 16 (a - 2) + 3 (10 - a) = 16 a - 32 + 30 - 3 a$ $rarr16(a-2) +3(10-a)=16a-32+30-3a$
Grouping the terms with unknowns and numbers to add/subtract them:
$16 a - 3 a - 32 + 30$ $16a-3a-32+30$
$= 13 a - 2$ $=13a-2$

Supposing that your question is to find the roots or find values of a such that $6 (a - 2) + 3 (10 - a) = 0$ $6(a-2) +3(10-a)=0$ :

$6 (a - 2) + 3 (10 - a) = 0$ $6(a-2) +3(10-a)=0$
$\to 13 a - 2 = 0$ $rarr 13a-2=0$

Adding 2 to both sides:
$13 a - 2 + 2 = 0 + 2$ $13a-2+2=0+2$
$13 a = 2$ $13a=2$

Dividing both sides by 13:
$13 \frac{a}{13} = \frac{2}{13}$ $13a/13=2/13$
$\to a = \frac{2}{13}$ $rarr a=2/13$

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Question #09738

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