Exercise 112 Page 562

We will solve the given system of equations using the Substitution Method. y=5x-22 & (I) y=x^2-6x+8 & (II) Notice that the y-variable is isolated in both equations. Since the expression equal to y in Equation (I) is simpler, we will substitute its value 5x-22 for y in Equation (II). Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. x^2-11x+30=0 ⇕ 1x^2+( - 11)x+ 30=0We can substitute a= 1, b= - 11, and c= 30 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 11)±sqrt(( - 11)^2-4( 1)( 30))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=11±sqrt((- 11)^2-4(1)(30))/2(1)

CalcPow

Calculate power

x=11±sqrt(121-4(1)(30))/2(1)

IdPropMult

Identity Property of Multiplication

x=11±sqrt(121-4(30))/2

Multiply

Multiply

x=11±sqrt(121-120)/2

SubTerm

Subtract term

x=11±sqrt(1)/2

CalcRoot

Calculate root

x=11± 1/2

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

x=11± 1/2
x_1=11+ 1/2	x_2=11- 1/2
x_1=12/2	x_2=10/2
x_1=6	x_2=5

Now, consider Equation (I). y=5x-22 We can substitute x=6 and x=5 into the above equation to find the values for y. Let's start with x=6.

y=5x-22

Substitute

x= 6

y=5( 6)-22

▼

Solve for y

Multiply

Multiply

y=30-22

SubTerm

Subtract term

y=8

We found that y=8 when x=6. One solution of the system, which is a point of intersection of the parabola and the line, is (6,8). To find the other solution we will substitute 5 for x in Equation (I) again.

y=5x-22

Substitute

x= 5

y=5( 5)-22

▼

Solve for y

Multiply

Multiply

y=25-22

SubTerm

Subtract term

y=3

We found that y=3 when x=5. Therefore our second solution, which is the other point of intersection of the parabola and the line, is (5,3).