Exercise 23 Page 246

We are asked to find a system of equations with (- 4, - 12) as its solution. For our first equation, we can choose any arbitrary slope and then substitute it, together with x= - 4 and y= - 12, in the general formula for the slope-intercept form of a line. Let's choose a slope of 2. We can find the second equation in the same way, as long as we do not choose the same slope! Let's now choose a slope of 5. Again, we know the line passes through ( - 4, - 12), and so this becomes our point.

y-y_1=m(x-x_1)

SubstituteValues

Substitute values

y-( -12)= 5(x-( -4))

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Solve for y

SubNeg

a-(- b)=a+b

y+12=5(x+4)

Distr

Distribute 5

y+12=5x+20

SubEqn

LHS-12=RHS-12

y=5x+8

When we combine these equations, we form a system of equations which has (- 4, - 12) as its solution. y=2x-4 y=5x+8