Exercise 21 Page 228

We are asked to find a system of equations with (3,5) as its solution. For our first equation, we can arbitrarily choose any slope and then substitute it, together with x= 3 and y= 5, into the general formula for a point-slope form of a line. y-y_1=m(x-x_1) Let's choose a slope of m= 2. y- 5= 2(x- 3) Now we can simplify this equation to be in slope-intercept form of a line, y=mx+b.

y- 5= 2(x- 3)

Distr

Distribute 2

y-5=2x-6

AddEqn

LHS+5=RHS+5

y=2x-1

We can find the second equation in the same way, as long as we do not choose the same value for the slope. Let's now choose a slope of 5. We know that this line also passes through ( 3, 5), so this becomes our point.

y-y_1=m(x-x_1)

SubstituteValues

Substitute values

y- 5= 5(x- 3)

Distr

Distribute 5

y-5=5x-15

AddEqn

LHS+5=RHS+5

y=5x-10

When we combine these equations, we form a system of equations which has (3,5) as its solution. y=2x-1 y=5x-10

Checking Our Answer

We can verify our answer by graphing the lines. Let's do it!