Page:Elementary algebra (1896).djvu/167

149 Rh To eliminate x we multiply (1) by 5 and (2) by 3, so as to make the coefficients of x in both equations equal. This gives

15x + 35y = 135, 15 x + 6 y = 48;

subtracting, 29 y = 87 ; y = 3.

To find x, substitute this value of y in either of the given equations.

Thus from (1) 3 x + 21 = 27 ; x= 2,, and y = 3.

In this solution we eliminated x by subtraction.

Rule. Multiply, when necessary, in such a manner as to make the coefficients of the unknown quantity to he eliminated equal in both equations. Add the resulting equations if these coefficients are unlike in sign; subtract if like in sign.

ELIMINATION BY SUBSTITUTION.

172. Ex. Solve 2x-5y=1 (1), 7x + 3y = 24 (2).

Transposing -5y in (1), and dividing by 2, we obtain x = {5y+1}{2 }

Substituting this value of x in (2) gives 7{5y+1}{2 } + 3y = 24 Whence 35y + 7 + 6 y = 48, and 41y = 41 ; y = 1.

This value substituted in either (1) or (2) gives x = 3.

Rule. From one of the equations, find the value of the unknown quantity to be eliminated in terms of the other and known quantities; then substitute this value for that quantity in the other equation, and reduce.