Looking for a geometry guru, help please

[majimikate]

Trying to solve for the sides of congruent triangle, which is congruent by SAS. The following 2 sides are congruent, AC=2x, and BD=x+5. I'm trying to solve for x...but I keep getting stuck...

2x=x+5

Subtract 5, get 2x-5=x
Divide by 2, get x=5/2????

Any help would be appreciated

 

[BCDisneyFanatic]

Trying to solve for the sides of congruent triangle, which is congruent by SAS. The following 2 sides are congruent, AC=2x, and BD=x+5. I'm trying to solve for x...but I keep getting stuck...

2x=x+5

Subtract 5, get 2x-5=x
Divide by 2, get x=5/2????

Any help would be appreciated

Subtract x from both sides:

2x-x = x-x +5

x = 5

 

[majimikate]

Thank you, but a quick question...since 2x is actually 2 times x, wouldn't you have to divide to x to isolate the 2, rather than be able to subtract it?

 

[kirstenb1]

I think the answer is 5

 

[BCDisneyFanatic]

Thank you, but a quick question...since 2x is actually 2 times x, wouldn't you have to divide to x to isolate the 2, rather than be able to subtract it?

You don't need to isolate the 2.

You first need to bring all of your variables to the same side of the equation.

You do this by subtracting the x from the right side of your equation. However, whatever you do to one side, you have to do to the other, so you subtract x from the lefthand side as well.

Now, instead of 2x = x+5

subtract x from both sides:

2x - x = (x-x) + 5

2x - x = 0 + 5

2x - x = 5

Now, looking only at the left hand side, think in terms of apples instead of x's.

2 apples - 1 apple
= 1 apple.

2 x - 1 x
= 1 x (or just x)

Therefore, we can replace the 2x-x with x

2x-x = 5
x = 5

Check:
left side: 2x 2(5) = 10
right side: x + 5 5 + 5 = 10

LS = RS therefore x = 5 works.

 

[BCDisneyFanatic]

2x=x+5

Subtract 5, get 2x-5=x
Divide by 2, get x=5/2????

If you were to divide 2x-5 = x by 2, you need to divide EVERYTHING by 2:

2x/2 - 5/2 = x/2

x - 5/2 = x/2

This doesn't really help you, though, since you still have variables on both sides of your equation.

You ALWAYS need to get all your variables to the same side of the equation. This is done by adding and subtracting. You would then divide or multiply both sides of the equation only if there is a coefficient in front of the variable.

For example, let's look at the equation:

3 x = x+10

Subtract x from both sides:

3x-x = x-x + 10

2x = 0 + 10

2x = 10

NOW, since there is the coefficient 2 in front of the x, you'd need to divide both sides by 2 to isolate x:

2x/2 = 10/2

x = 5