solve the following pairs of linear equations 5x+3y-8=0;3x-2y+75=0
Answers
Answer:
5x+3y-8=0 and 3x-2y +75 = 0. Now let us consider 5x+3y-8 as equation (1) and 3x-2y+75=0
as equation (2). Now multiply equation (1) by 2, we get (5x+3y-8)(2)=0(2)= 10x+6y-16=0. Now multiply equation (2) by 3, we get (3x-2y+75)(3)
=0(3). i.e 9x-6y+225=0. Now we got 2 equations,
10x+6y-16=0 and 9x-6y+225=0. By adding the 2 equations, we get (10x+6y-16=0)+(9x-6y+225=0).
We get 19x +209=0 19x= -209, x= -209/19, x=-11.
Hence we got x=-11, By substituting it in the equation 1, we get 5x+3y-8=0, 5(-11)+3y -8=0, -55+3y-8=0, -63+3y=0, 3y=63, y=21. Hence we get x=-11 and y=21. Hope this helps you.
Step-by-step explanation:
You may get 1 doubt, why we are multiplying equation (1) by 2 and equation(2) by 3, Because by multiplying we get 6y in equation 1 and -6y in equation 2, Hence by adding them , we get -6y+6y=0, Hence we get the value of x, and by substituting it in either equation 1 or 2, we get the value of y. Hence we got the values of x and y. Hope this helps you.