Question #09738

2 Answers
Sep 11, 2017

color(blue)(13a - 2 )13a2

Explanation:

16( a - 2 ) + 3( 10 - a )16(a2)+3(10a)

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

16a - 32 + 30 -3a16a32+303a

Collect like terms:

16a -3a -32 + 30 = color(blue)(13a -2) 16a3a32+30=13a2

Sep 11, 2017

=13a-2=13a2

Explanation:

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

rarr16(a-2) +3(10-a)=16a-32+30-3a16(a2)+3(10a)=16a32+303a
Grouping the terms with unknowns and numbers to add/subtract them:
16a-3a-32+3016a3a32+30
=13a-2=13a2

Supposing that your question is to find the roots or find values of a such that 6(a-2) +3(10-a)=06(a2)+3(10a)=0:

6(a-2) +3(10-a)=06(a2)+3(10a)=0
rarr 13a-2=013a2=0

Adding 2 to both sides:
13a-2+2=0+213a2+2=0+2
13a=213a=2

Dividing both sides by 13:
13a/13=2/1313a13=213
rarr a=2/13a=213