SOLUTION: Solve each system by using both the substitution and addition methods: A. 4x-y=6 x-y=0 B. 5x+2y=1 x-3y=7

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Question 89000: Solve each system by using both the substitution and addition methods:
A. 4x-y=6
x-y=0
B. 5x+2y=1
x-3y=7
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A.

Here is the elimination method:
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

4%2Ax-1%2Ay=6
1%2Ax-1%2Ay=0

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 4 and 1 to some equal number, we could try to get them to the LCM.

Since the LCM of 4 and 1 is 4, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -4 like this:

1%2A%284%2Ax-1%2Ay%29=%286%29%2A1 Multiply the top equation (both sides) by 1
-4%2A%281%2Ax-1%2Ay%29=%280%29%2A-4 Multiply the bottom equation (both sides) by -4


So after multiplying we get this:
4%2Ax-1%2Ay=6
-4%2Ax%2B4%2Ay=0

Notice how 4 and -4 add to zero (ie 4%2B-4=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%284%2Ax-4%2Ax%29-1%2Ay%2B4%2Ay%29=6%2B0

%284-4%29%2Ax-1%2B4%29y=6%2B0

cross%284%2B-4%29%2Ax%2B%28-1%2B4%29%2Ay=6%2B0 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

3%2Ay=6

y=6%2F3 Divide both sides by 3 to solve for y



y=2 Reduce


Now plug this answer into the top equation 4%2Ax-1%2Ay=6 to solve for x

4%2Ax-1%282%29=6 Plug in y=2


4%2Ax-2=6 Multiply



4%2Ax=6%2B2 Subtract -2 from both sides

4%2Ax=8 Combine the terms on the right side

cross%28%281%2F4%29%284%29%29%2Ax=%288%29%281%2F4%29 Multiply both sides by 1%2F4. This will cancel out 4 on the left side.


x=2 Multiply the terms on the right side


So our answer is

x=2, y=2

which also looks like

(2, 2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

4%2Ax-1%2Ay=6
1%2Ax-1%2Ay=0

we get



graph of 4%2Ax-1%2Ay=6 (red) 1%2Ax-1%2Ay=0 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (2,2). This verifies our answer.


------------------------------------------------
Here is the substitution method:
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

4%2Ax-1%2Ay=6
1%2Ax-1%2Ay=0

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y.

Solve for y for the first equation

-1%2Ay=6-4%2AxSubtract 4%2Ax from both sides

y=%286-4%2Ax%29%2F-1 Divide both sides by -1.


Which breaks down and reduces to



y=-6%2B4%2Ax Now we've fully isolated y

Since y equals -6%2B4%2Ax we can substitute the expression -6%2B4%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


1%2Ax%2B-1%2Ahighlight%28%28-6%2B4%2Ax%29%29=0 Replace y with -6%2B4%2Ax. Since this eliminates y, we can now solve for x.

1%2Ax-1%2A%28-6%29-1%284%29x=0 Distribute -1 to -6%2B4%2Ax

1%2Ax%2B6-4%2Ax=0 Multiply



1%2Ax%2B6-4%2Ax=0 Reduce any fractions

1%2Ax-4%2Ax=0-6 Subtract 6 from both sides


1%2Ax-4%2Ax=-6 Combine the terms on the right side



-3%2Ax=-6 Now combine the terms on the left side.


cross%28%281%2F-3%29%28-3%2F1%29%29x=%28-6%2F1%29%281%2F-3%29 Multiply both sides by 1%2F-3. This will cancel out -3%2F1 and isolate x

So when we multiply -6%2F1 and 1%2F-3 (and simplify) we get



x=2 <---------------------------------One answer

Now that we know that x=2, lets substitute that in for x to solve for y

1%282%29-1%2Ay=0 Plug in x=2 into the 2nd equation

2-1%2Ay=0 Multiply

-1%2Ay=0-2Subtract 2 from both sides

-1%2Ay=-2 Combine the terms on the right side

cross%28%281%2F-1%29%28-1%29%29%2Ay=%28-2%2F1%29%281%2F-1%29 Multiply both sides by 1%2F-1. This will cancel out -1 on the left side.

y=-2%2F-1 Multiply the terms on the right side


y=2 Reduce


So this is the other answer


y=2<---------------------------------Other answer


So our solution is

x=2 and y=2

which can also look like

(2,2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

4%2Ax-1%2Ay=6
1%2Ax-1%2Ay=0

we get


graph of 4%2Ax-1%2Ay=6 (red) and 1%2Ax-1%2Ay=0 (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle.


and we can see that the two equations intersect at (2,2). This verifies our answer.


-----------------------------------------------------------------------------------------------
Check:

Plug in (2,2) into the system of equations


Let x=2 and y=2. Now plug those values into the equation 4%2Ax-1%2Ay=6

4%2A%282%29-1%2A%282%29=6 Plug in x=2 and y=2


8-2=6 Multiply


6=6 Add


6=6 Reduce. Since this equation is true the solution works.


So the solution (2,2) satisfies 4%2Ax-1%2Ay=6



Let x=2 and y=2. Now plug those values into the equation 1%2Ax-1%2Ay=0

1%2A%282%29-1%2A%282%29=0 Plug in x=2 and y=2


2-2=0 Multiply


0=0 Add


0=0 Reduce. Since this equation is true the solution works.


So the solution (2,2) satisfies 1%2Ax-1%2Ay=0


Since the solution (2,2) satisfies the system of equations


4%2Ax-1%2Ay=6
1%2Ax-1%2Ay=0


this verifies our answer.





===============================================

B.


Here is the elimination method:
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

5%2Ax%2B2%2Ay=1
1%2Ax-3%2Ay=7

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 5 and 1 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and 1 is 5, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -5 like this:

1%2A%285%2Ax%2B2%2Ay%29=%281%29%2A1 Multiply the top equation (both sides) by 1
-5%2A%281%2Ax-3%2Ay%29=%287%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
5%2Ax%2B2%2Ay=1
-5%2Ax%2B15%2Ay=-35

Notice how 5 and -5 add to zero (ie 5%2B-5=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%285%2Ax-5%2Ax%29%2B%282%2Ay%2B15%2Ay%29=1-35

%285-5%29%2Ax%2B%282%2B15%29y=1-35

cross%285%2B-5%29%2Ax%2B%282%2B15%29%2Ay=1-35 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

17%2Ay=-34

y=-34%2F17 Divide both sides by 17 to solve for y



y=-2 Reduce


Now plug this answer into the top equation 5%2Ax%2B2%2Ay=1 to solve for x

5%2Ax%2B2%28-2%29=1 Plug in y=-2


5%2Ax-4=1 Multiply



5%2Ax=1%2B4 Subtract -4 from both sides

5%2Ax=5 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%285%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=1 Multiply the terms on the right side


So our answer is

x=1, y=-2

which also looks like

(1, -2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax%2B2%2Ay=1
1%2Ax-3%2Ay=7

we get



graph of 5%2Ax%2B2%2Ay=1 (red) 1%2Ax-3%2Ay=7 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (1,-2). This verifies our answer.


------------------------------------------------
Here is the substitution method:
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

5%2Ax%2B2%2Ay=1
1%2Ax-3%2Ay=7

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y.

Solve for y for the first equation

2%2Ay=1-5%2AxSubtract 5%2Ax from both sides

y=%281-5%2Ax%29%2F2 Divide both sides by 2.


Which breaks down and reduces to



y=1%2F2-%285%2F2%29%2Ax Now we've fully isolated y

Since y equals 1%2F2-%285%2F2%29%2Ax we can substitute the expression 1%2F2-%285%2F2%29%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


1%2Ax%2B-3%2Ahighlight%28%281%2F2-%285%2F2%29%2Ax%29%29=7 Replace y with 1%2F2-%285%2F2%29%2Ax. Since this eliminates y, we can now solve for x.

1%2Ax-3%2A%281%2F2%29-3%28-5%2F2%29x=7 Distribute -3 to 1%2F2-%285%2F2%29%2Ax

1%2Ax-3%2F2%2B%2815%2F2%29%2Ax=7 Multiply



1%2Ax-3%2F2%2B%2815%2F2%29%2Ax=7 Reduce any fractions

1%2Ax%2B%2815%2F2%29%2Ax=7%2B3%2F2Add 3%2F2 to both sides


1%2Ax%2B%2815%2F2%29%2Ax=14%2F2%2B3%2F2 Make 7 into a fraction with a denominator of 2


1%2Ax%2B%2815%2F2%29%2Ax=17%2F2 Combine the terms on the right side



%282%2F2%29%2Ax%2B%2815%2F2%29x=17%2F2 Make 1 into a fraction with a denominator of 2

%2817%2F2%29%2Ax=17%2F2 Now combine the terms on the left side.


cross%28%282%2F17%29%2817%2F2%29%29x=%2817%2F2%29%282%2F17%29 Multiply both sides by 2%2F17. This will cancel out 17%2F2 and isolate x

So when we multiply 17%2F2 and 2%2F17 (and simplify) we get



x=1 <---------------------------------One answer

Now that we know that x=1, lets substitute that in for x to solve for y

1%281%29-3%2Ay=7 Plug in x=1 into the 2nd equation

1-3%2Ay=7 Multiply

-3%2Ay=7-1Subtract 1 from both sides

-3%2Ay=6 Combine the terms on the right side

cross%28%281%2F-3%29%28-3%29%29%2Ay=%286%2F1%29%281%2F-3%29 Multiply both sides by 1%2F-3. This will cancel out -3 on the left side.

y=6%2F-3 Multiply the terms on the right side


y=-2 Reduce


So this is the other answer


y=-2<---------------------------------Other answer


So our solution is

x=1 and y=-2

which can also look like

(1,-2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax%2B2%2Ay=1
1%2Ax-3%2Ay=7

we get


graph of 5%2Ax%2B2%2Ay=1 (red) and 1%2Ax-3%2Ay=7 (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle.


and we can see that the two equations intersect at (1,-2). This verifies our answer.


-----------------------------------------------------------------------------------------------
Check:

Plug in (1,-2) into the system of equations


Let x=1 and y=-2. Now plug those values into the equation 5%2Ax%2B2%2Ay=1

5%2A%281%29%2B2%2A%28-2%29=1 Plug in x=1 and y=-2


5-4=1 Multiply


1=1 Add


1=1 Reduce. Since this equation is true the solution works.


So the solution (1,-2) satisfies 5%2Ax%2B2%2Ay=1



Let x=1 and y=-2. Now plug those values into the equation 1%2Ax-3%2Ay=7

1%2A%281%29-3%2A%28-2%29=7 Plug in x=1 and y=-2


1%2B6=7 Multiply


7=7 Add


7=7 Reduce. Since this equation is true the solution works.


So the solution (1,-2) satisfies 1%2Ax-3%2Ay=7


Since the solution (1,-2) satisfies the system of equations


5%2Ax%2B2%2Ay=1
1%2Ax-3%2Ay=7


this verifies our answer.