Subtitution and Elimination #2

Question

Solve each system of equations by using the substitution method
1.  y=x+5

     y=4x-1

 

2.  y=x/2

     y = x-4

 

3.  x+y=5

     2x-3y=5

Answer 1

-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

 

      

 

Answer 2

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.  

   

Answer 3

Hi Andrea;

1. y=x+5
y=4x-1

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!

2. y=x/2
y = x-4
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!

3. x+y=5
2x-3y=5

 

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.