Exercise 2 Page 212

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. -5x+2 y=13 & (I) 5x+ y=-1 & (II) We can see that the x-terms will eliminate each other if we add Equation (II) to Equation (I). Now, we can solve for x by substituting the value of y into either equation and simplifying.

y=4 5x+y=-1

Substitute

(II): y= 4

y=4 5x+ 4=-1

▼

(II):Solve for x

SubEqn

(II): LHS-4=RHS-4

y=4 5x+4-4=-1-4

SubTerms

(II): Subtract terms

y=4 5x=-5

DivEqn

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

y=4 5x/5=-5/5

MovePartNumRight

(II): a* b/c=a/c* b

y=4 5/5 * x=-5/5

MoveNegNumToFrac

(II): Put minus sign in front of fraction

y=4 5/5 * x=-5/5

QuotOne

(II): a/a=1

y=4 1 * x=-1

IdPropMult

(II): Identity Property of Multiplication

y=4 x=-1

The solution, or point of intersection, of the system of equations is (-1,4). To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct.

-5x+2y=13 5x+y=-1

(I), (II): x= -1, y= 4

-5( -1)+2( 4) ? = 13 5( -1)+ 4 ? = -1

(I), (II): Multiplication Property of -1

5+2(4) ? = 13 -5+4 ? = -1

Multiply

(I): Multiply

5+8 ? = 13 -5+4 ? = -1

(I), (II): Add and subtract terms

13 = 13 ✓ -1 = -1 ✓