1. y=x+5

Solve each system of equations by using the substitution method

1. y=x+5

1. y=x+5

y=4x-1

2. y=x/2

y = x-4

3. x+y=5

2x-3y=5

Tutors, sign in to answer this question.

-5x+10y=60

-4x+16y=32

the idea is to eliminate one of the variables; there are many ways to do this

multiply both sides of the first equation by 4 and multiply both sides of the second equation by 5

-20x+40y=240

-20x+80y=160

subtract the second equation from the first equation

-40y=80 (-20x-(-20x))=0, eliminating x

y=-2

substitute y=-2 into either equation(the new ones or the original ones)

-20x+(40)(-2)=240

-20x-80=240

-20x=320

x=-16

x=-16, y=-2

if you substituted into the second equation you would get:

-20x+(80)(-2)=160

-20x-160=160

-20x=320

x=-16 again

using the original equations: -5x+10(-2)=60

-5x-20=60

-5x=80

x=-16

-4x+16(-2)=32

-4x-32=32

-4x=64

x=-16

check: -5x+10y=60

(-5)(-16)+(10)(-2)=60

80+(-20)=60

60=60

-4x+16y=32

(-4)(-16)+(16)(-2)=32

64+(-32)=32

32=32

(-1/2)x-(1/4)y=3

(1/3)x+(1/6)y=2

again, there are many ways to solve this problem

multiply both sides of the first equation by 6 and multiply both sides of the second equation by 9

(6)(-1/2)x-(6)(1/4)y=6(3)

(9)(1/3)x+(9)(1/6)y=9(2)

we get -3x-1.5y=18

3x+1.5y=18

now add the two equations: 0+0=36

0=36

this is a false statement, therefore there is no solution

another solution would be to multiply both sides of both equations by the LCM of 2,3,4, and 6 which is 12 and then work with the new equations without any fractions

Hi Andrea;

y=y

Let's substitute...

x+5=4x-1

Let's isolate x...

Let's subtract 5 from both sides...

-5+x+5=4x-1-5

x=4x-6

Let's subtract 4x from both sides...

-4x+x=4x-6-4x

-3x=-6

Let's divide both sides by -3...

(-3x)/-3=-6/-3

x=2

Let's plug the value of x into the first equation to establish the value of y...

y=x+5

y=2+5

y=7

Let's plug the value of x into the second equation to establish the value of y and verify the results of the first equation...

y=4x-1

y=4(2)-1

y=8-1

y=7

Results match!

y=y

Let's substitute...

x/2=x-4

Let's eliminate the denominator, also known as the divisor, of 2 by dividing both sides by 2...

2(x/2)=2(x-4)

x=2x-8

Let's subtract 2x from both sides as we proceed to isolate x...

-2x+x=2x-8-2x

-x=-8

Let's multiply both sides by -1...

(-1)(-x)=-8(-1)

x=8

Let's plug this value of x into the first equation and solve for y...

y=x/2

y=8/2

y=4

Let's plug this value of x into the second equation and solve for y to verify the results of the first...

y=x-4

y=8-4=4

The results match!

Let's isolate y using either equation. The first is the easiest...

x+y=5

y=5-x

2x-3y=5

Let's substitute y for (5-x)...

(2x)-[3(5-x)]=5

2x-15+3x=5

Let's combine like terms...

5x-15=5

Let's add 15 to both sides as we proceed to isolate x...

15+5x-15=5+15

5x=20

Let's divide both sides by 5...

(5x)/5=20/5

x=4

Let's plug this value of x into the first equation to solve for y...

x+y=5

4+y=5

y=1

Let's plug this value of x into the second equation to solve for y and verify the results of the first equation...

2x-3y=5

2(4)-3y=5

8-3y=5

-3y=-3

y=1

The results match.

Hi. typo - Sorry. 1) y=x+5

y=4x-1

Hi Andrea;

y=x+5

y=4x-1

y=y

Let's substitute...

x+5=4x-1

Do you need additional help?

Hi Vivien! Thank you so much. I THINK I'm getting the hang of it. I'm going to try a few on my own and hopefully I won't have to post for help. lol

Andrea;

Post all of the help you want. I enjoy it. I just want to make certain that I am not enjoying it so much that the help is harmful.

Hi Vivian. I've been really stuck on these two equations and I know how much you like them. I do understand now how to do the substituion, but these two are not the same as the others. Can you help?

-5x+10x=60

-4x+16y=32

and

-1/2x-1/4y=3

1/3x+1/6y=2

Hi Andrea;

Thank you for complimenting me with your request for help. I appreciate it.

I just arrived home from work. Can you give me a few minutes?

Hi Andrea;

I was thinking about you at work, wondering how you are doing. I saw your second request last night, then went to sleep. I have been doing too many work-hours recently.

The first equation must have a typo in it as only one variable, x, is identified. I assume there is a y.

As to the second set of equations, I have struggled with these for over one hour. I cannot figure it out. Is there another typo?

Hi Andrea;

I TRIED EVERYTHING AND CANNOT SOLVE THIS.

THIS IS WHAT I HAVE.

CAN YOU TALK TO OTHER STUDENTS AND ASK IF THEY ARE HAVING THE SAME ISSUE?

-1/2x-1/4y=3

I am assuming this is...

(-1/2)x-(1/4)y=3

Let's eliminate the fractions by multiply both sides by 4...

4[(-1/2)x-(1/4)y]=3(4)

-2x-y=12

1/3x+1/6y=2

Let's eliminate the fractions by multiplying both sides by 6...

6[(1/3)x+(1/6)y]=2(6)

2x+y=12

Let's isolate y...

Let's subtract 2x from both sides...

-2x+2x+y=12-2x

y=12-2x

First equation...

-2x-y=12

Let's substitute 12-2x for y...

-2x-(12-2x)=12

-4x-12=12

Let's add 12 to both sides...

12-4x-12=12+12

-4x=24

Let's divide both sides by -4...

(-4x)/-4=24/-4

x=-6

Back to the original equations...

Let's solve for y...

-1/2x-1/4y=3

[(-1/2)(-6)]-(1/4)y=3

3-[(1/4)y]=3

Let's subtract 3 from both sides...

-3+3-[(1/4)y]=3-3

-(1/4)y=0

y=0

1/3x+1/6y=2

(1/3)(-6)+(1/6)y=2

-2+[(1/6)y]=2

Let's add 2 to both sides...

2-2+[(1/6)y]=2+2

(1/6)y=4

Let's multiply both sides by 6...

6(1/6)y=4(6)

y=24

It does not work.

I can't solve it.

Nancy K. | A-1 Experienced Resource Teacher - All Grades A-1 Experienced Resource Teacher - All G...

1. In this example, we are told that y=x+5. Therefore, we can substitute x+5 for the y in the second equation. However, since what should be the second equation is missing the =, it is not possible to solve it.

2. In this equation, we are told that y=x/2. Therefore, we can substitute x/2 for y in the second equation.

x/2=x-4

The next step is to multiply the entire equation by 2 in order to eliminate the fraction.

2(x/2) = 2(x-4)

After simplifying, we have x=2x-8.

Subtract 2x from both sides and you have -x = -8. Therefore, **x=8**. When we substitute 8 for x in the first equation, we see that
**y = 4.** We confirm that by inserting both values into the second equation.

Since 4=8-4 is a true statement, the solution is correct.

3. This set of equations can also be solved by substitution. First, since x+y=5, then x=5-y (subtract y from both sides. Now substitute 5-y for x in the second equation.

2(5-y) -3y = 5

Distribute the 2 and get 10-2y-3y=5. Combine the like terms and get 10-5y=5. Subtract 10 from both sides and get -5y = -5. Therefore,
**y=1**.

Substitute 1 for y in the first equation. x+1=5. Subtract 1 from both sides;
**x=4.** When we insert both values into the second equation, we get 8-3=5. Since that is a true statement, our solution is correct.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments