Substitution Method?

Question

I have three questions that I need help with
3x+y=5
-2x+3y=4

y=5x
2x+y=28

x+3y=23
3x-2y=3


How do I use the substitution method with these problems

Answer 1

Using the substitution method, you solve for one variable in terms of the other.
For example, 3x+y = 5 means that y = 5-3x
So, in the SECOND equation, everywhere you see a "y" you put in 5-3x instead:
-2x + 3(5-3x) = 4
and then solve this equation for x:
-2x + 15 - 9x = 4
-11x = -11
x = 1
then substitute back into the "y = 5-3x" equation to get y = 5 - 3*1 = 2

For the second one, it's SIMPLE: substituts 5x for y in the second equation and see what you get.

For the third, it's easier, since x is all by itself, to say x = 23 - 3y
and in the second equation, 3(23-3y) - 2y = 3 you can solve for y.
Give it a shot. You won't learn this if I do the whole thing!

Answer 2

You substitute an equation for a variable from one of the two equations into the other equation
say A + B = 10
B + C = 100
1. pick a variable (in my example you can ONLY pick B because it is the only variable occuring in both equations)
so, B it is.
2. Then pick an equation - lets pick the first. It doesn't matter too much.
3 .Solve for B in this equation - easy (?) subtract A from both sides
B = 10 - A
So we have A+B=10 and finally we get to the substitution
4. we will susbtitute 10-A for B in the OTHER equation!!! THis trips a lot of people up - do not substitute into the same (first) equation or you'll go round in circles!
So the second equation is B + C = 100
substituting we get 10 - A + C = 100
This is the substitution method. In my example above it didn't help. At All!
But lets try another with two equations and just two unkowns.
A + 2B = 3
2A + 3B = 6
1 .Pick B, again
2. Lets do the second equation first this time.
2A + 3B = 6
3. Solve for the variable you picked (B)
subtract 2A from both sides
3B = 6 - 2A
divide by 3
B = 2 - (2/3)A
4. Substitute
A + 2B = 3 = A + 2{ 2 - (2/3)A} = 3
expand
A + 4 - 4/3A = 3
whats A - 4/3A ?? = (-1/3)A
so..... (-1/3)A + 4 = 3
subtract 4 from both sides
(-1/3)A = -1
multiply both sides by - 3
A = 3
Almost there!
Substitute 3 for A in either equation (do the first its easier)
A + 2B = 3 = 3 +2B oh, oh B is Zero!
does it work for the other equation too? 2*3 + 3*0 = 6 right?
oup A=3, B=0 tricky. Don't forget to
5. substitute the variable you have solved for and get the other variable too. SO it takes two substitutions if you have two variables.
6. Finally check that the numbers you got are right by substituting them back into both equations (when you get to be a pro at it, you'll only have to check one equation)

Answer 3

In each pair solve one of the equations for x then plug what you get into the second equation.
For the first pair I will solve 3x + y = 5 for x
3x + y = 5
3x = 5-y
x = (5-y)/3
plug this x into the second equation
-2((5-y)/3) + 3y = 4
(-10+2y)/3 + 3y = 4
-10/3 + 2y/3 + 3y = 4
-10/3 +11y/3 = 4
11y/3 = 4 +10/3
11y/3 = 23/3
y = 23/11
now find x by plugging the y you found into one of the equations.
I will put it into the first equation
3x + 23/11 = 5
3x = 5 - 23/11
3x = 32/11
x = 32/33

Another way to do the problems is to find y and plug that into the second equation. Do that for the second pair since they tell you y = 5x.

Answer 4

Taking the first set as an example, start by solving the first equation for y. You get y = -3x + 5. Then use this value for y in the second equation. -2x + 3y = 4 becomes -2x + 3(-3x + 5) = 4. This expands to -2x - 9x + 15 = 4, and simplifies to -11x = -11, a simple algebraic problem. When you get the value of x, you can subsitute it into either of the original equations and solve algebraically for y.

Answer 5

1)3x+y=5,-2x+3y=4
Subtract 3x from both sides:
y=-3x+5

Put it into this equation:
-2x+3(-3x+5)=4
-2x-9x+15=4
-11x+15=4
-11x=-11
x=1

y=-3(1)+5
y=-3+5
y=2

The solution set is (1,2)

Check:
3(1)+2=5
3+2=5
5=5

-2(1)+3(2)=4
-2+6=4
4=4

2)y=5x,2x+y=28
Just take y=5x and substitute it into the next equation:
2x+5x=28
7x=28
x=4

y=5(4)
y=20

The solution set is (4,20)

Check:
20=5(4)
20=20

2(4)+20=28
8+20=28
28=28

3)x+3y=23,3x-2y=3
Subtract 3y from both sides:
x=-3y+23

Now, substitute that into the second equation:
3(-3y+23)-2y=3
-9y+69-2y=3
-11y+69=3
-11y=-66
y=6

x+3(6)=23
x+18=23
x=5

The solution set is (5,6)

Check:
5+3(6)=23
5+18=23
23=23

3(5)-2(6)=3
15-12=3
3=3

Answer 6

1.y=5-3x
-2x+3(5-3x)=4
-2x+15-9x=4
-11x+15-15=4-15
-11x=-11
x=1
sub y=2

2.2x+5x=28
7x=28
7x/7=28/7
x=4

3.x=23-3y
3(23-3y)-2y=3
69-9y-2y=3
-11y+69-69=3-69
-11y=-66
-11y/-11=-66/-11
y=6
sub x=5