# 13x - 5y = 10 3x - 2y = 14

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Solve the simultaneous equations 13x-5y=10; 3x-2y=14:

We can use linear combinations (multiplication/addition method);

13x-5y=10
3x-2y=14       Multiply eq1 by -2 and eq2 by 5

-26x+10y=-20
15x-10y=70   Add the two equations:

-11x+0y=50

`x=-50/11`

Substitute this into one of the original equations:

`3(-50/11)-2y=14`

`-2y=304/11`

`y=-152/11`

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The solution is `(-50/11,-152/11)`

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Check: `13(-50/11)-5(-152/11)=-650/11+760/11=110/11=10`

`3(-50/11)-2(-152/11)=-150/11+304/11=154/11=14`

jess1999 | Student, Grade 9 | (Level 1) Valedictorian

Posted on

13x - 5y = 10

3x - 2y = 14

First multiply everything in the first equation by 2 and everything in the second equation by 5 .

By multiplying , you should get

26x - 10y = 20

15x - 10y = 70  now subtract the two equation .

By subtracting , you should get

11x = -50 now divide both sides by 11 .

By dividing both sides by 11 , you should get

x = -50/11 which is your answer for " x "

Now plug -50/11 into one of the equation as " x "

3 ( -50/11 ) - 2y = 14 multiply

-150/11 - 2y = 14 add

-2y = 304/11 now divide by -2 on both sides

y = -152/11  which is your answer for " y "

So " x " is -50/11 and " y " is -152/11

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`13x - 5y = 10` --- (i)

`3x - 2y = 14`   ---- (ii)

These equations have to be solved.

Solution of the system of equations through elimination of y requires the following operation:

(i)*2-(ii)*5

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`26x - 10y = 20`

`15x - 10y = 70`

(-)--------------------------

`11x=-50`

`x=-50/11`

Put this value of x in eq.(i) to get,

`13*-50/11-2y=10`

`rArr 2y=(-650/11-10)=-760/11`

`rArr y=-380/11`

Therefore, the solution of the above equations is:

`x=-50/11`

`y=-380/11`