Solving Using Substitution

**annon25** · November 29th 2011, 12:31 PM

Hey Everyone! I'm going to have a few emergency help questions today and early tomorrow. I have a huge test.

My problem right now is solving with substitution, and I am given the following problem to do so with:

4x+y=5
2x-3y=13

I'm thinking I should bring up the 2x or the 3 from the bottom equation to the top? T

Thanks!

**Quacky** · November 29th 2011, 12:35 PM

Rearranging the first to isolate $y$ :

$y=5-4x$
Now attempt to substitute this into the second equation. If you can't, then please show your working.

**pickslides** · November 29th 2011, 01:18 PM

Multiply the second equation through by 2 then subtract the result from the first equation.

**Soroban** · November 29th 2011, 01:26 PM

Hello, annon25!

Solve by substitution: . $\begin{array}{cccc}4x+y&=&5 & [1] \\ 2x-3y&=&13 & [2]\end{array}$

I'm thinking I should bring up the 2x or the 3 from the bottom equation to the top?
What does that mean?

Do you understand the substitution method?

[1] Solve one of the equations for one of its variables.
[2] Substitute this into the other equation and solve for the remaining variable.
[3] Then solve for the other variable.

Solve [1] for $y\!:\;\;y \:=\:5-4x$

Substitute into [2]: . $2x - 3(5-4x) \:=\:13$

. . . . and we have: . $2x - 15 + 12x \:=\:13 \quad\Rightarrow\quad 14x \:=\:28 \quad\Rightarrow\quad x \,=\,2$

Substitute into [1]: . $4(2) + y \:=\:5 \quad\Rightarrow\quad y \,=\,\text{-}3$

Therefore: . $\begin{Bmatrix}x &=& \;2 \\ y &=& \text{-}3 \end{Bmatrix}$

**e^(i*pi)** · November 29th 2011, 01:28 PM

Originally Posted by annon25

Hey Everyone! I'm going to have a few emergency help questions today and early tomorrow. I have a huge test.

My problem right now is solving with substitution, and I am given the following problem to do so with:

4x+y=5
2x-3y=13

I'm thinking I should bring up the 2x or the 3 from the bottom equation to the top? T

Thanks!

Since the questions states substitution I would not use pickslide's methods (which is elimination). What it wants you to do is sub an expression for either x or y into the other equation solving for one of the variables then finding the other.

In your case it is easier to subtract 4x from both sides to get $y = 5-4x$ . You can now sub 5-4x for y in the second equation: $2x-3(5-4x) = 13$ .

You can now find x and then use the first equation to find y

**annon25** · November 29th 2011, 01:32 PM

Okay,
This is what I ended up with:

2x-4x-3y

If I plug in 2 as x and-3 as y, I am getting the answer "5..."

**e^(i*pi)** · November 29th 2011, 01:35 PM

Originally Posted by annon25

Okay,
This is what I ended up with:

2x-4x-3y

If I plug in 2 as x and-3 as y, I am getting the answer "5..."

Soroban has answered your question.

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