Exercise 40 Page 180

1. Isolate a variable in one of the equations.
2. Substitute the expression for that variable into the other equation and solve.
3. Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate y in the first equation.

Now that we've isolated y, we can solve the system by substitution.

Solving this system of equations resulted in an identity; 14 is always equal to itself. Therefore, the lines are the same and have infinitely many intersection points.

y=2(x+3)^2-5 & (I) y=14x+17 & (II) Since the expression equal to y in (II) is simpler, let's use that for our initial substitution.

Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable.

x^2-x-2=0 ⇔ 1x^2+( -1)x+( - 2)=0 Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= -1, and c= -2 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

▼

Solve for x

NegNeg

- (- a)=a

x=1±sqrt((-1)^2-4(1)(-2))/2(1)

CalcPow

Calculate power

x=1±sqrt(1-4(1)(-2))/2(1)

MultByOne

a * 1=a

x=1±sqrt(1-4(-2))/2

MultNegNegOnePar

- a(- b)=a* b

x=1±sqrt(1+8)/2

AddTerms

Add terms

x=1±sqrt(9)/2

CalcRoot

Calculate root

x=1± 3/2

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=1± 3/2
x_1=1+3/2	x_2=1-3/2
x_1=4/2	x_2=-2/2
x_1=2	x_2=-1

Now, consider Equation (II). y=14x+17 We can substitute x=2 and x=-1 into the above equation to find the values for y. Let's start with x=2.

y=14x+17

Substitute

x= 2

y=14( 2)+17

▼

Solve for y

Multiply

Multiply

y=28+17

AddTerms

Add terms

y=45

We found that y=45 when x=2. One solution of the system, which is a point of intersection of the parabola and the line, is (2,45). To find the other solution, we will substitute -1 for x in Equation (II) again.

y=14x+17

Substitute

x= -1

y=14( -1)+17

▼

Solve for y

MultPosNeg

a(- b)=- a * b

y=-14+17

AddTerms

Add terms

y=3

We found that y=3 when x=-1. Therefore our second solution, which is the other point of intersection, is ( -1,3). Therefore the graphs of the given equations intersect at two points.

y=3(x-2)^2+3 & (I) y=6x-12 & (II) Since the expression equal to y in (II) is simpler, let's use that for our initial substitution.

Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable.

x^2-6x+9=0 ⇔ 1x^2+( -6)x+ 9=0 Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= -6, and c= 9 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=-( -6)±sqrt(( -6)^2-4( 1)( 9))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=6±sqrt((-6)^2-4(1)(9))/2(1)

CalcPow

Calculate power

x=6±sqrt(36-4(1)(9))/2(1)

MultByOne

a * 1=a

x=6±sqrt(36-4(9))/2

MultNegPos

(- a)b = - ab

x=6±sqrt(36-36)/2

SubTerm

Subtract term

x=6±sqrt(0)/2

CalcRoot

Calculate root

x=6/2

CalcQuot

Calculate quotient

x=3

This result tells us that we have one solution for x. We can substitute x=3 into the second equation to find the value for y.

y=6x-12

Substitute

x= 3

y=6( 3)-12

▼

Solve for y

Multiply

Multiply

y=18-12

SubTerm

Subtract term

y=6

We found that y=6, when x=3. Therefore the graphs of given equations intersect at one point.