# Thread: Solve by substitution method

1. ## Solve by substitution method

Hello i need help on this problem

Solve by substitution
x+y=6
y=x+2

2. Originally Posted by volcombro2000
Hello i need help on this problem

Solve by substitution
x+y=6
y=x+2
As it's name suggests, when we solve by substitution, we substitute the expression for one variable in one equation for that variable in the other equation. from the second equation, we see that y = x + 2, so we simply substitute x + 2 for y in the other eqaution. we would then have an equation with only x, which is easy to solve.

$\displaystyle x + y = 6 ...........(1)$
$\displaystyle y = x + 2 ...........(2)$

From (2) we see $\displaystyle y = x + 2$
substitute $\displaystyle x + 2$ for $\displaystyle y$ in (1), we get:

$\displaystyle x + (x + 2) = 6$

$\displaystyle \Rightarrow 2x + 2 = 6$

$\displaystyle \Rightarrow 2x = 4$

$\displaystyle \Rightarrow x = 2$

Now, can you tell me what $\displaystyle y$ is?

3. y=4

4. ## Is Elmination a simlar concept

Solve by Elmination

x+y=15
6x-y+41

5. Originally Posted by volcombro2000
Solve by Elmination

x+y=15
6x-y+41
no, elimination is a different concept. here we try to ELIMINATE one variable from one of our equations by adding or subtracting the two equations. Sometimes we may have to multiply one or both equations by a constant so we can eliminate a variable. here we see that we have a y in one equation and a -y in the other. if we add the two equations then, we would get rid of the y, since y + (-y) = 0y

$\displaystyle x + y = 15 ............(1)$
$\displaystyle 6x - y = 41 ..........(2)$

$\displaystyle \Rightarrow 7x = 56 ............(1)+(2)$

$\displaystyle \Rightarrow x = 8$

Can you tell me what $\displaystyle y$ is?

Originally Posted by volcombro2000
y=4
Correct!