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Subtitution and Elimination #2

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3 Answers

-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
 
      
 

Comments

I did the problem three different ways and each time I got a false statement, therefore there is no solution.
Also, put both equations in the slope-intercept form and you get y=-2x-12 and y=-2x+12; both equations have the same slope and therefore are parallel and don't intersect, thus no solution.
Thank you Arthur!!!!!!!!!!!!!
 
I forgot that one detail.  But will never again forget.  Amazing how we remember our mistakes more than our successes!
 
If the two equations are parallel, these cannot share a coordinate.  I struggled so much with these equations, and thought about it today at work.  I was worried about Andrea going to school confused.
 
Andrea, please confirm you are fine!
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.

Comments

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.
 
 
 
 
 
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.