# Solve by substitution method

• June 1st 2007, 03:36 PM
volcombro2000
Solve by substitution method
Hello i need help on this problem

Solve by substitution
x+y=6
y=x+2
• June 1st 2007, 03:44 PM
Jhevon
Quote:

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.

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

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

$x + (x + 2) = 6$

$\Rightarrow 2x + 2 = 6$

$\Rightarrow 2x = 4$

$\Rightarrow x = 2$

Now, can you tell me what $y$ is?
• June 1st 2007, 04:10 PM
volcombro2000
y=4
• June 1st 2007, 04:13 PM
volcombro2000
Is Elmination a simlar concept
Solve by Elmination

x+y=15
6x-y+41
• June 1st 2007, 04:34 PM
Jhevon
Quote:

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

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

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

$\Rightarrow x = 8$

Can you tell me what $y$ is?

Quote:

Originally Posted by volcombro2000
y=4

Correct! :D