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

Algebra ->  Linear-equations -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

a.
4x-3y=10 ….this is written in the standard form
y+-20=-6x….write this one in the standard form too
6x+%2B+y+=20….
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

4%2Ax-3%2Ay=10
6%2Ax%2B1%2Ay=20

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:

3%2A%284%2Ax-3%2Ay%29=%2810%29%2A3 Multiply the top equation (both sides) by 3
-2%2A%286%2Ax%2B1%2Ay%29=%2820%29%2A-2 Multiply the bottom equation (both sides) by -2


So after multiplying we get this:
12%2Ax-9%2Ay=30
-12%2Ax-2%2Ay=-40

Notice how 12 and -12 add to zero (ie 12%2B-12=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2812%2Ax-12%2Ax%29-9%2Ay-2%2Ay%29=30-40

%2812-12%29%2Ax-9-2%29y=30-40

cross%2812%2B-12%29%2Ax%2B%28-9-2%29%2Ay=30-40 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:

-11%2Ay=-10

y=-10%2F-11 Divide both sides by -11 to solve for y



y=10%2F11 Reduce


Now plug this answer into the top equation 4%2Ax-3%2Ay=10 to solve for x

4%2Ax-3%2810%2F11%29=10 Plug in y=10%2F11


4%2Ax-30%2F11=10 Multiply



4%2Ax-30%2F11=10 Reduce



4%2Ax=10%2B30%2F11 Subtract -30%2F11 from both sides

4%2Ax=110%2F11%2B30%2F11 Make 10 into a fraction with a denominator of 11

4%2Ax=140%2F11 Combine the terms on the right side

cross%28%281%2F4%29%284%29%29%2Ax=%28140%2F11%29%281%2F4%29 Multiply both sides by 1%2F4. This will cancel out 4 on the left side.


x=35%2F11 Multiply the terms on the right side


So our answer is

x=35%2F11, y=10%2F11

which also looks like

(35%2F11, 10%2F11)

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

4%2Ax-3%2Ay=10
6%2Ax%2B1%2Ay=20

we get



graph of 4%2Ax-3%2Ay=10 (red) 6%2Ax%2B1%2Ay=20 (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 (35%2F11,10%2F11). This verifies our answer.



b.
+y=+3x+%2B+6+…. write this one in the standard form
-3x+%2B+y=+6+….
+6x+%2B+2y+=+12…. 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

-3%2Ax%2B1%2Ay=6
6%2Ax%2B2%2Ay=12

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%2B3%2AxAdd 3%2Ax to both sides

y=%286%2B3%2Ax%29 Divide both sides by 1.


Which breaks down and reduces to



y=6%2B3%2Ax Now we've fully isolated y

Since y equals 6%2B3%2Ax we can substitute the expression 6%2B3%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


6%2Ax%2B2%2Ahighlight%28%286%2B3%2Ax%29%29=12 Replace y with 6%2B3%2Ax. Since this eliminates y, we can now solve for x.

6%2Ax%2B2%2A%286%29%2B2%283%29x=12 Distribute 2 to 6%2B3%2Ax

6%2Ax%2B12%2B6%2Ax=12 Multiply



6%2Ax%2B12%2B6%2Ax=12 Reduce any fractions

6%2Ax%2B6%2Ax=12-12 Subtract 12 from both sides


6%2Ax%2B6%2Ax=0 Combine the terms on the right side



12%2Ax=0 Now combine the terms on the left side.


cross%28%281%2F12%29%2812%2F1%29%29x=%280%2F1%29%281%2F12%29 Multiply both sides by 1%2F12. This will cancel out 12%2F1 and isolate x

So when we multiply 0%2F1 and 1%2F12 (and simplify) we get



x=0 <---------------------------------One answer

Now that we know that x=0, lets substitute that in for x to solve for y

6%280%29%2B2%2Ay=12 Plug in x=0 into the 2nd equation

0%2B2%2Ay=12 Multiply

2%2Ay=12%2B0Add 0 to both sides

2%2Ay=12 Combine the terms on the right side

cross%28%281%2F2%29%282%29%29%2Ay=%2812%2F1%29%281%2F2%29 Multiply both sides by 1%2F2. This will cancel out 2 on the left side.

y=12%2F2 Multiply the terms on the right side


y=6 Reduce


So this is the other answer


y=6<---------------------------------Other answer


So our solution is

x=0 and y=6

which can also look like

(0,6)

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

-3%2Ax%2B1%2Ay=6
6%2Ax%2B2%2Ay=12

we get


graph of -3%2Ax%2B1%2Ay=6 (red) and 6%2Ax%2B2%2Ay=12 (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 (0,6). This verifies our answer.


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

Plug in (0,6) into the system of equations


Let x=0 and y=6. Now plug those values into the equation -3%2Ax%2B1%2Ay=6

-3%2A%280%29%2B1%2A%286%29=6 Plug in x=0 and y=6


0%2B6=6 Multiply


6=6 Add


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


So the solution (0,6) satisfies -3%2Ax%2B1%2Ay=6



Let x=0 and y=6. Now plug those values into the equation 6%2Ax%2B2%2Ay=12

6%2A%280%29%2B2%2A%286%29=12 Plug in x=0 and y=6


0%2B12=12 Multiply


12=12 Add


12=12 Reduce. Since this equation is true the solution works.


So the solution (0,6) satisfies 6%2Ax%2B2%2Ay=12


Since the solution (0,6) satisfies the system of equations


-3%2Ax%2B1%2Ay=6
6%2Ax%2B2%2Ay=12


this verifies our answer.