Solve the system of equation {5x+6y=4 and 3x+7y=8 by the method of elimination

Question

Solve the system of equation {5x+6y=4 and 3x+7y=8 by the method of elimination

Answer 1

Hi Grayson.

 

Elimination is the process where you "eliminate" one of the variables.  Sometimes it is easy, like if they give you

 

2x  - 3y = 8

5x + 3y = 6

 

If you were to add these veritcally,  you would get 2x + 5x = 7x   and -3y + 3y = 0  (our goal) and 8 + 6 = 14

 

Now you have 7x = 14,  so x = 2.  Plug this back into either original equation and get that y = -(4/3)

 

 

Back to your problem.

 

Neither of these variables will eliminate easily.

 

So you will have to FORCE one of the variable to "go away".

 

5x + 6y = 4

3x + 7y = 8

 

To make "x" go away, need to multiply entire top equation  by (-3) and the whole bottom equation by (5).

 

What this does is give a -15x on top and a +15x on the bottom, so we can now eliminate x.

 

-15x - 18y = -12

 15x + 35y = 40

 

Add these vertically and get 

 

0x + 17y = 28

17y = 28

y = (28/17)

 

Could plug this back in, but is a mess to work with fractions, so just go back and now eliminate "y" by multiplying the entire top equation by (-7) and the entire bottom equation by 6.

 

-35x - 42y = -28

 18x + 42y = 48

 

Adding vertically:

 

-17x + 0y = 20

 

-17x = 20

 

x = -(20/17)

 

 

And you can check your answer by substituting x = -(20/17) and y = (28/17)  back into BOTH original equations and make sure they work in both original equations.

 

5(-20/17) + 6(28/17) = 4

-100/17 + 168/17 = 68/17 = 4  so this solution works in the first equation

 

3(-20/17) + 7(28/17) = 8

-60/17 + 196/17 = 136/17 = 8  so this solution also works in the 2nd equation, so it must be correct!

Answer 2

5x+6y=4

3x+7y=8

 

In elimination method,we have to get rid off either of the variable x or y.

Let us try to eliminate x

To get rid off x,we have to make coefficient of x in both equations as same.

So,we multiply first equation with coefficient of x from second equation,this means we multiply first equation by 3

and we multiply second equation with coefficient of x from first  equation,this means we multiply second equation by 5

(5x+6y=4)*3 ----->>>  15x+18y=12
(3x+7y=8)*5 ------>>> 15x+35y=40

Next we subtract these equations

15x+18y=12
15x+35y=40

-    -        -

-----------------

 0x  -17y=-28   

        

divide both sides by -17 

We get y=28/17

 

Next we need to find value of x

For this we plug in value of y=28/17 into either of original equation

Let us plug in it first equation

5x+6(28/17) = 4

5x+168/17 = 4

5x= 4  - 168/17

5x= 68/17  -168/17

5x= -100/17

Divide both sides by 5

x= -100/(5*17)

x=-100/85

We reduce this fraction by 5

x=-20/17

 

Finally we get x=-20/17 and y=28/17

Answer 3

The goal here is to eliminate either one of your variables, so lets say you wish to eliminate  x in order to solve for y.

 

Firstly, you'll need to multiply 5x +6y = 4 by (-3) and 3x +7y = 8 by (5), which will yield their equivalent counterparts with coefficients that cancel out:

 

 

     5x +6y = 4   (-3)   ---->   -15x -18y = -12

 

     3x +7y = 8   (5)    ---->    15x +35y = 40

 

 

 

Secondly, you can now add the obtained equations and solve for y:

 

 

     15x +35y = 40

 

    -15x -18y = -12

------------------------

       0  +17y = 28          ---->     y = 28/17

 

 

 

Finally, you can now use the obtained value for y to solve for x, so plug the y value into either one of the original equations, say 5x +6y = 4 and solve it for x:

 

 

  5x +6(28/17) = 4     ---->    5x +(168/17) = 4    ---->    5x = 4 -(168/17)    ---->   

 

    5x = -(100/17)

 

 

 

Dividing both sides by 5 should yield x = -20/17 ... and you're done!!