# Solve the following equations for x and y: 8x + 3y = 1 and 6x + 2y = 4

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We have to solve the equations: 8x + 3y = 1 and 6x + 2y = 4 for x and y.

8x + 3y = 1… (1)

6x + 2y = 4 … (2)

We now solve by performing 2*(1) – 3*(2)

=> 2*(8x + 3y) – 3*(6x + 2y) = 2*1 – 3*4

=> 16x + 6y – 18x – 6y = 2 – 12

=> -2x = -10

=> x = 5

Now substitute x = 5 in (1)

=> 8*5 + 3y = 1

=> 40 + 3y = 1

=> 3y = -39

=> y = -13

**Therefore x = 5 and y = -13**

8x+3y=1............eq (1)

6x+2y=4.............eq (2)

**multiplying eq(1) with 6 and eq (2) with 8**

so we get 48x+18y=6

48x+16y=32

(-) (-) (-)

__________

2y=-26

y=-26/2 =-13 so **y=-13**

**putting the value of y in eq (1)**

8x+3y=1

8x+3(-13)=1

8x-39=1 ** (minus sign is greater so we put minus sign)**

8x=1+39** (minus is converted into plus) **

8x=40

x=40/8

**x=5**

*x=5 y=-13*

8x+3y = 1...(1).

6x+2y = 4....(2).

We are required to solve the above sytem of equations:

We divide the equation 6x+2y = 4 by 2 and we get:

3x+y = 2. Therefore y = 2-3x. We substitute y= 3x in equation 8x+3y and we get: 8x+3(2-3x) = 1.Or 8x+6-9x = 1. 8x-9x = 1-6 = -5. Or -x= -5.

So x = 5.

We substitute x = 5 in equation 8x+3y = 1. So 8*5 +3y = 1. Or 3y = 1-40 = -39. So 3y= -39. Or y = -39/3 = -13.

So y = -13.

Therefore x = 5 and y = -13 are the solutions of the given system of equations.

We'll solve th system using elimination method.

We notice that the second equation of the system could be divided by 2:

6x + 2y = 4

3x + y = 2 (2)

8x + 3y = 1 (1)

We'll multiply by -3 the equation (2):

-9x -3y = -6 (3)

We'll add (3)+(1):

-9x -3y + 8x + 3y = -6 + 1

We'll combine and eliminate like terms:

-x = -5

**x = 5**

We'll substitute x in (1):

40 + 3y = 1

3y = 1 - 40

3y = -39

**y = -13**

**The solution of the system is: {5 , -13}.**