please help me solve this problems the long way to help me understand each steps.

**BeBeMala** · March 29th 2010, 08:43 PM

if 3x+4x+35=21, then find the value of 2x+10?

a) 6
B)-2
c)-4
d)-6

a triangle with perimeter 40 has a short side x, a middle side of 7 less than twice the shorter and a long side of 7 more than the middle side. which equation can be used to solve for x?

a) 4x=40
b) 5x-7=40
c)4x-7=40
d)5x-14=40

**Anonymous1** · March 29th 2010, 09:05 PM

Well...

$3x+4x+35=21$

$7x + 35 = 21$ Combine like terms

7 $x = 21 - 35 = -14$ Subtract $35$ from both sides

$x= \frac{-14}{7} = -2$ Divide both sides by $7$

So $2x+10$ is,

$2(-2) + 10 = 6$

**Anonymous1** · March 29th 2010, 09:10 PM

Perimeter = small side + middle side + long side

$40 = [x] + [2x-7] + [7+(2x-7)]$

You finish this one...

**BeBeMala** · March 29th 2010, 09:18 PM

Originally Posted by Anonymous1

Perimeter = small side + middle side + long side

$40 = [x] + [2x-7] + [7+(2x-7)]$

You finish this one...

i need to combine life term?
for example 5x-7=40 is the answer?

**Anonymous1** · March 29th 2010, 09:27 PM

Correct. Do you understand now?

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