Exercise 24 Page 246

We are asked to find a system of equations with (15,- 25) as its solution. For our first equation, we can choose any arbitrary slope and then substitute it, together with x= 15 and y= - 25, 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 3. Again, we know the line passes through ( 15, - 25), and so this becomes our point.

y-y_1=m(x-x_1)

SubstituteValues

Substitute values

y-( -25)= 3(x- 15)

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Solve for y

SubNeg

a-(- b)=a+b

y+25=3(x-15)

Distr

Distribute 3

y+25=3x-45

SubEqn

LHS-25=RHS-25

y=3x-70

When we combine these equations, we form a system of equations which has (15, -25) as its solution. y=2x-55 y=3x-70