Exercise 37 Page 248

To begin, if x=- 3 is in fact a solution to the equation it means we can substitute x=- 3 into the equation and it will make a true statement. mx+b=- 2x-1 To determine the necessary values of m and b, we will hence begin by substituting x=- 3 into the equation and simplify.

mx+b=- 2x-1

Substitute

x= - 3

m( - 3)+b=- 2( - 3)-1

Multiply

Multiply

- 3m +b = 6-1

SubTerms

Subtract terms

- 3m +b =5

At this point, our equation is - 3m+b=5. We can now arbitrarily choose the value of one of the variables. For example, if we let m=1 we can find the corresponding value of b.

- 3m+b=5

Substitute

m= 1

- 3* 1+b=5

MultByOne

a * 1=a

- 3 +b =5

AddEqn

LHS+3=RHS+3

b=8

Thus, if m=1 then b=8. Let's write our equation with those values. x+8=- 2x-1 By graphing the lines y=x+8 and y=- 2x-1 we can make sure that the x-coordinate of the intersection point is x=- 3.

The lines intersect at (- 3,5), which means the solution to the equation is in fact x=- 3.