Exercise 37 Page 594

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. This means that either the x-terms or the y-terms must cancel each other out. -2 x+15 y=-32 & (I) 7 x-5 y=17 & (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 (II) by 3, the y-terms will have opposite coefficients. -2 x+15 y=-32 3(7 x-5 y)=3(17) ⇒ -2 x+ 15y=-32 21 x- 15y=51 We can see that the y-terms will eliminate each other if we add (I) to (II).

-2 x+15y=-32 21x-15y=51

SysEqnAdd

(II): Add (I)

-2 x+15y=-32 21x-15y+( - 2x+15y)=51+( -32)

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(II):Solve for x

AddSubTerms

(II): Add and subtract terms

-2 x+15y=-32 19x=19

DivEqn

(II): .LHS /19.=.RHS /19.

-2 x+15y=-32 x=1

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

-2 x+15y=-32 x=1

Substitute

(I): x= 1

-2 ( 1)+15y=-32 x=1

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(I):Solve for y

MultByOne

(I): a * 1=a

-2+15y=-32 x=1

AddEqn

(I): LHS+2=RHS+2

15y=-30 x=1

DivEqn

(I): .LHS /15.=.RHS /15.

y=-2 x=1

The solution, or point of intersection, of the system of equations is (1,-2).