Math HELP!!!!!!!!!?

Question

solve each equation using the algebraic method that seems convenient
3a - b =13
2a + 3b = 16

i asked a question similar to this but i still dont really get it

Answer 1

Multiply the first equation by 3 to make the b coefficients
the same.
So 9a -3b = 39
and 2a + 3b = 16
Now if you add the 2 equations, the b terms cancel.
So 11a = 55
a = 5
3(5) - b = 13
b = 2.
The solution is a = 5, b = 2.
If you still have troubles, please e-mail me and
I'll try to help you more.

Answer 2

3a - b = 13
2a + 3b = 16

You will use one of two methods:
1) substitution, or
2) elimination.

I will show you both methods.

1) By substitution:
In one equation, express one variable in terms of the other, and then plug into the second equation.

3a - b = 13
2a + 3b = 16

Let's solve for b in the first equation.
3a - b = 13
b = 3a - 13

And now we plug this into the second equation.

2a + 3(3a - 13) = 16
2a + 9a - 39 = 16
11a - 39 = 16
11a = 55
a = 5

Now that you know a, you can solve for b. Plug the value a = 5 into either equation to get b. But since we've already solved b = 3a - 13, then b = 3(5) - 13 = 15 - 13 = 2
a = 5, b = 2

2) By elimination:

Multiply one of the equations by the appropriate number so that you may add the equations to eliminate one variable.

3a - b = 13
2a + 3b = 16

Multiply the first equation by 3,

9a - 3b = 39
2a + 3b = 16

Add the equations together. Note that -3b and +3b add to 0, effectively cancelling each other out.

9a + 2a = 39 + 16
11a = 55
a = 5

We can use the same method as above to get b. Plugging in a = 5 into either equation:

3a - b = 13
3(5) - b = 13
15 - b = 13
15 - 13 = b
2 = b, or b = 2

a = 5, b = 2

Answer 3

When you have two equations each with two unknown numbers (like x or y or some letter) you can solve them at the same time to find the answers

This is called Solving Equations Simultaneously.

You have to fiddle with the equations until one unknown (such as a or b in this case) cancels each other out.

3a - b = 13
2a + 3b = 16

To make one cancel the other you have to have one equal the other except have one as a postive and one as a negative. Just as -3 and +3 cancel each other out so would something like -4b and +4b, same idea.

So you can choose which unknown you wish to get rid of. I'll choose b:

3a - b = 13
2a + 3b = 16

If I want -b to cancel with +3b I'll have to make them both into 3's or both into 1's (-b is -1b you just don't write the 1). Seeing as it's easier to go up I've decided to make them both into 3's

So you have one +3b and one -b, they both have opposite signs so I don't need to bother changing that. But what do I have to do to -b to make it -3b so it will cancel??

Multiply by 3 right? (-b x 3 = -3b)

So in order to multiply -b by 3 I have to multiply the whole equation it's in by 3 to keep it true (As you know you can't just do something to one number and not to the rest in the equation)

So multiply (3a - b = 13) by 3:

9a - 3b = 39 (I've multiplied each number by 3)

Now we're ready to simultaniously solve them:

9a - 3b = 39
2a + 3b = 16

Basically conjoin them and add them downwards:

9a -3b = 39
2a + +3b+ = 16+
___ ___ ___
11a 0 = 55

As you see the b's have cancelled and we're left with 11a = 55

Solve this and we find x:

11a = 55
a = 5

Hard part over yay

Now to find b substitute your new value for a (5) into any of the two equations. I'll choose the first:


3(5) - b = 13 (I've replaced the a's with 5's)
15 - b = 13
- b = -2 (Reverse the signs)
b = 2

Therfore your answers are:
a = 5
b = 2

Hope that helps!

Answer 4

3a - b =13
2a + 3b = 16

1. You need to multiply the first equation by 3 so that you can get rid of "b".

9a = 39
2a = 16

2. Then you can add the like terms.

11a = 55

3. Then solve for a.

a = 5

4. To find b, just plug in the value of "a" to the equation of your choice.

You answer would be:

a = 5
b = 2

Hope this helps:)!