SOLUTION: solve one using addition and one using substitution. State which method your are using for which system a. 4x-3y=10 and y -20=-6x b. y= 3x + 6 and =6x + 2y = 12 Thanks

Question 120266: solve one using addition and one using substitution.
State which method your are using for which system
a. 4x-3y=10 and y -20=-6x
b. y= 3x + 6 and =6x + 2y = 12
Thanks
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

a.
�.this is written in the standard form
�.write this one in the standard form too
�.
Here is solution (using addition):

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition

Lets start with the given system of linear equations

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 6 to some equal number, we could try to get them to the LCM.

Since the LCM of 4 and 6 is 12, we need to multiply both sides of the top equation by 3 and multiply both sides of the bottom equation by -2 like this:

Multiply the top equation (both sides) by 3
Multiply the bottom equation (both sides) by -2

So after multiplying we get this:

Notice how 12 and -12 add to zero (ie )

Now add the equations together. In order to add 2 equations, group like terms and combine them

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:

Divide both sides by to solve for y

Reduce

Now plug this answer into the top equation to solve for x

Plug in

Multiply

Reduce

Subtract from both sides

Make 10 into a fraction with a denominator of 11

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.

Multiply the terms on the right side

So our answer is

,

which also looks like

(, )

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

we get

graph of (red) (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 (,). This verifies our answer.

b.
�. write this one in the standard form
�.
�. this is the standard form
Here is solution (using substitution):

Solved by pluggable solver: Solving a linear system of equations by subsitution

Lets start with the given system of linear equations

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

Add to both sides

Divide both sides by 1.

Which breaks down and reduces to

Now we've fully isolated y

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

Replace y with . Since this eliminates y, we can now solve for x.

Distribute 2 to

Multiply

Reduce any fractions

Subtract from both sides

Combine the terms on the right side

Now combine the terms on the left side.

Multiply both sides by . This will cancel out and isolate x

So when we multiply and (and simplify) we get

<---------------------------------One answer

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

Plug in into the 2nd equation

Multiply

Add to both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out 2 on the left side.

Multiply the terms on the right side

Reduce

So this is the other answer

<---------------------------------Other answer

So our solution is

and

which can also look like

(,)

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

we get

graph of (red) and (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 (,). This verifies our answer.

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

Plug in (,) into the system of equations

Let and . Now plug those values into the equation

Plug in and

Multiply

Add

Reduce. Since this equation is true the solution works.

So the solution (,) satisfies

Let and . Now plug those values into the equation

Plug in and

Multiply

Add

Reduce. Since this equation is true the solution works.

So the solution (,) satisfies

Since the solution (,) satisfies the system of equations

this verifies our answer.

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