Exercise 117 Page 494

We want to solve the given system.

{\frac{x}{4} + \frac{y}{3} = 1 2 x - \frac{y}{3} = 17 (I) (II)

Before finding the method to solve this, let's simplify the equations. We can get rid of the fractions by multiplying each equation by a common denominator of the fractions in it. Let's do it!

{\frac{x}{4} + \frac{y}{3} = 1 2 x - \frac{y}{3} = 17 (I) (II)

MultEqn

$(I):$ $LHS \cdot 12 = RHS \cdot 12$

{3 x + 4 y = 12 2 x - \frac{y}{3} = 17

MultEqn

$(II):$ $LHS \cdot 3 = RHS \cdot 3$

{3 x + 4 y = 12 6 x - y = 51

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other. This means that either the

x -

or the

y -

terms must cancel each other out.

{3 x + 4 y = 12 6 x - y = 51 (I) (II)

Currently, none of the terms in this system will cancel out. Let's find a common multiple between two variable like terms in the system. If we multiply Equation (I) by

- 2,

the

x -

terms will have opposite coefficients.