SOLUTION: i dont kno0w how to sove for substitution elimination or graphing for this problem 2x-y=1 and 3x+y=-6

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Question 120350: i dont kno0w how to sove for substitution elimination or graphing for this problem 2x-y=1 and 3x+y=-6
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

2%2Ax-1%2Ay=1
3%2Ax%2B1%2Ay=-6

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

y=%281-2%2Ax%29%2F-1 Divide both sides by -1.


Which breaks down and reduces to



y=-1%2B2%2Ax Now we've fully isolated y

Since y equals -1%2B2%2Ax we can substitute the expression -1%2B2%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


3%2Ax%2B1%2Ahighlight%28%28-1%2B2%2Ax%29%29=-6 Replace y with -1%2B2%2Ax. Since this eliminates y, we can now solve for x.

3%2Ax%2B1%2A%28-1%29%2B1%282%29x=-6 Distribute 1 to -1%2B2%2Ax

3%2Ax-1%2B2%2Ax=-6 Multiply



3%2Ax-1%2B2%2Ax=-6 Reduce any fractions

3%2Ax%2B2%2Ax=-6%2B1Add 1 to both sides


3%2Ax%2B2%2Ax=-5 Combine the terms on the right side



5%2Ax=-5 Now combine the terms on the left side.


cross%28%281%2F5%29%285%2F1%29%29x=%28-5%2F1%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5%2F1 and isolate x

So when we multiply -5%2F1 and 1%2F5 (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

3%28-1%29%2B1%2Ay=-6 Plug in x=-1 into the 2nd equation

-3%2B1%2Ay=-6 Multiply

1%2Ay=-6%2B3Add 3 to both sides

1%2Ay=-3 Combine the terms on the right side

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

y=-3%2F1 Multiply the terms on the right side


y=-3 Reduce


So this is the other answer


y=-3<---------------------------------Other answer


So our solution is

x=-1 and y=-3

which can also look like

(-1,-3)

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

2%2Ax-1%2Ay=1
3%2Ax%2B1%2Ay=-6

we get


graph of 2%2Ax-1%2Ay=1 (red) and 3%2Ax%2B1%2Ay=-6 (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,-3). This verifies our answer.


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Check:

Plug in (-1,-3) into the system of equations


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

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


-2%2B3=1 Multiply


1=1 Add


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


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



Let x=-1 and y=-3. Now plug those values into the equation 3%2Ax%2B1%2Ay=-6

3%2A%28-1%29%2B1%2A%28-3%29=-6 Plug in x=-1 and y=-3


-3-3=-6 Multiply


-6=-6 Add


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


So the solution (-1,-3) satisfies 3%2Ax%2B1%2Ay=-6


Since the solution (-1,-3) satisfies the system of equations


2%2Ax-1%2Ay=1
3%2Ax%2B1%2Ay=-6


this verifies our answer.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Substitution:

Eq. 1) 2x-y=1
Eq. 2) 3x%2By=-6

Solve equation 1 for y:

2x-y=1
-y=-2x%2B1
y=2x-1

Take the expression just developed for y and substitute it into equation 2 in place of the y:

3x%2By=-6
3x%2B%282x-1%29=-6

Now solve for x

3x%2B%282x-1%29=-6
5x-1=-6
5x=-6%2B1
5x=-5
x=-1}

Substitute this value for x into equation 1 and solve for y:

2%28-1%29-y=1
-2-y=1
-y=1%2B2
y=-3

Therefore the solution set is the ordered pair (-1,-3)

Elimination:


Eq. 1) 2x-y=1
Eq. 2) 3x%2By=-6

Add the two equations term-by-term:

Sum: 5x=-5

Solve for x

x=-1

Since you have a -y in one equation and a y in the other, you can add the two equations directly to eliminate the y variable. In other problems where the coefficients are not direct additive inverses on one of the variables, you will have to multiply through one of the equations by an appropriate value to create the additive inverse situation.

To illustrate the process, I'll solve for the y variable by elimination of the x variable:
2x-y=1
3x%2By=-6

Multiply equation 1 by 3, and equation 2 by -2:

6x-3y=3
-6x-2y=12

Now add the equations term by term:
Sum: -5y=15

And solve for y:

y=-3

Again, the solution set is the ordered pair (-1,-3) (And I sincerely hope you aren't surprised that the two methods produce the same result. I wouldn't have been surprised necessarily if they had been different, but I would have been supremely confident that I had made an error in the performance of one or both of the methods.)

Graphing:

Select two x values for each of the lines, and then compute the corresponding y value for each selected x. This will give you two ordered pairs for each line. Plot the points and construct a straight line through each set of points. The solution will be where the two lines intersect.

graph%28400%2C400%2C-5%2C5%2C-5%2C5%2C2x-1%2C-3x-6%29

And, as we suspected, the point of intersection is (-1,-3)

Hope this helps,
John