Exercise 59 Page 64

Since neither equation in the given system has a variable with a coefficient of 1, we will use the Elimination Method. To do this, 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. - 5 x+2 y=13 & (I) 2 x+3 y=- 9 & (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 (I) by 3 and multiply (II) by -2, the y-terms will have opposite coefficients. 3(- 5 x+2 y)= 3(13) -2(2 x+3 y)= -2(-9) ⇓ - 15 x+ 6y=39 - 4 x- 6y=18 We can see that the y-terms will eliminate each other if we add (I) to (II).

- 15x+6y=39 - 4x-6y=18

SysEqnAdd

(II): Add (I)

- 15x+6y=39 - 4x-6y+( - 15x+6y)=18+ 39

▼

(II):Solve for x

RemovePar

(II): Remove parentheses

- 15x+6y=39 - 4x-6y-15x+6y=18+39

AddSubTerms

(II): Add and subtract terms

- 15x+6y=39 - 19x=57

DivEqn

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

- 15x+6y=39 x=- 3

Now we can solve for y by substituting the value of x into (I) and simplifying.

- 15x+6y=39 x=- 3

Substitute

(I): x= - 3

- 15( - 3)+6y=39 x=- 3

▼

(I):Solve for y

MultNegNegOnePar

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

45+6y=39 x=- 3

SubEqn

(I): LHS-45=RHS-45

6y=- 6 x=- 3

DivEqn

(I): .LHS /6.=.RHS /6.

y=- 1 x=- 3

The solution, or point of intersection, of the system of equations is (- 3, - 1).