can anyone solve this algebra 1 problem?

5(2x+7)=42

1. Start off by distributing the 5. You do this by multiplying it by 2x and 7.

So you get this:

5*2x=10x

5*7=35

Which means:

10x+35=42

2. Next, you subtract 35 from both sides of the equal sign.

So you get this:

10x+35=42

-35=-35

Which equals

10x=7

3. Next, divide by 10 on each side.

4: So you get

x=7/10

Hope this helps! :D

* = Multiplication, by the way.

5(2x+7)=42

Distribute the 5 (5*2x and 5*7)

10x+35=42

Next subtract 35 from both sides (35-35 and 42-35)

10x=7

now you need x by its self so divide by 10 (10x/10 and 7/10)

which equals .7 or you can leave it as 7/10 if your teacher will let you

then plug it in and it works

Good luck :)

You need to distribute so it would be

10x+35=42 then you subtract 35 from both sides then dived by 10 so the full equation is

10x+35=42

-35 -35

10x=7

___ __

10 10

X=7/10

10x+35=42

42-35=7

10x=7

x=10/7

10x+35= 42

10x = 7

x = 7/10

5 (2x + 7)= 42

10x + 35 = 42

-35 -35

10x =7

----- ----

10 10

x = .7 or 7/10

x=((42/50)-5)/2