If a+b=81 and b+c=23, find 3a+8b+5c

358

Step-by-step explanation:

If we multiply the first equation by 3 on both sides we get 3a+3b=243.

If we multiply the second equation by 5 on both sides we get 5b+5c=115.

Adding these equations gives

(3a+3b)+(5b+5c)=243+115

Combine like terms:

3a+8b+5c=358

Another way.

Find numbers to satisfy the criteria.

a+b=81

b+c=23

Choose a=80.

This forces b=1 since 80+1=81.

If b=1 and b+c=23, this forces c=22.

Now plug into 3a+8b+5c.

3(80)+8(1)+5(22)

240+8+110

358

128

Step-by-step explanation:

a+b =81

a =81-b

b+c= 23

c =23-b

3a+8b+5c

= 3(81-b) +8b+5(23-b)

=243-3b+8b+ 115-5b

=128+5b-5b

=128

Notice: I'm not really sure.

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

answer: 48

just remember to use pemdas:

parentheses

expressions

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