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

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

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Saugus, MA

-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

Middletown, CT

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.

Miami, FL

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.

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