Exercise 14 Page 391

Since neither equation has a variable with a coefficient of 1, the Substitution Method may not be the easiest. Instead, we will use the Elimination Method. To solve a system using this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This means that either the x-terms or the y-terms must cancel each other out. 4 x+5 y=3 & (I) 3 x-2 y=8 & (II) Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply Equation (I) by 3 and multiply Equation (II) by -4, the x-terms will have opposite coefficients. 3(4 x+5 y)=3(3) -4(3 x-2 y)=-4(-8) ⇒ 12x+15 y=9 - 12x+8 y=32We can see that the x-terms will eliminate each other if we add Equation (I) to Equation (II).

12x+15y=9 & (I) -12x+8y=-32 & (II)

SysEqnAdd

(II): Add I

12x+15y=9 -12x+8y+ 12x+15y=-32+ 9

▼

(II): Solve for y

AddSubTerms

(II): Add and subtract terms

12x+15y=9 23y=-23

DivEqn

(II): .LHS /23.=.RHS /23.

12x+15y=9 y=-1

Now, we can solve for x by substituting the value of y into either equation and simplifying.

12x+15y=9 y=-1

Substitute

(I): y= -1

12x+15( -1)=9 y=-1

▼

(I): Solve for x

MultPosNeg

(I):a(- b)=- a * b

12x-15=9 y=-1

AddEqn

(I):LHS+15=RHS+15

12x=24 y=-1

DivEqn

(I):.LHS /12.=.RHS /12.

x=2 y=-1

The solution to this system of equations is (2,-1).