Exercise 9 Page 393

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 equation. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. 2 x+ 5y=2 & (I) 3 x- 5y=53 & (II) We can see that the y-terms in this system will eliminate each other if we add (I) to (II).

2x+5y=2 & (I) 3x-5y=53 & (II)

SysEqnAdd

(II): Add (I)

2x+5y=2 3x-5y+ 2x+5y=53+ 2

▼

(II):Solve for x

AddSubTerms

(II): Add and subtract terms

2x+5y=2 5x=55

DivEqn

(II): .LHS /5.=.RHS /5.

2x+5y=2 x=11

We can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation.

2x+5y=2 x=11

Substitute

(I): x= 11

2( 11)+5y=2 x=11

▼

(I):Solve for y

Multiply

(I): Multiply

22+5y=2 x=11

SubEqn

(I): LHS-22=RHS-22

5y=-20 x=11

DivEqn

(I): .LHS /5.=.RHS /5.

y=-4 x=11

The solution, or point of intersection, of the system of equations is (11,-4).