Exercise 30 Page 722

We want to solve the given system of equations using the Substitution Method. y=x^2-x-6 & (I) y=2x-2 & (II) The y-variable is isolated on both equations. Let's substitute the right-hand side of Equation (II) for y into Equation (I).

y=x^2-x-6 y=2x-2

Substitute

(I): y= 2x-2

2x-2=x^2-x-6 y=2x-2

▼

(I): Simplify

SubEqn

(I): LHS-2x=RHS-2x

- 2=x^2-3x-6 y=2x-2

AddEqn

(I): LHS+2=RHS+2

0=x^2-3x-4 y=2x-2

RearrangeEqn

(I): Rearrange equation

x^2-3x-4=0 y=2x-2

Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2-3x-4=0 ⇕ 1x^2+( - 3)x+( - 4)=0We can substitute a= 1, b= - 3, and c= - 4 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 3)±sqrt(( - 3)^2-4( 1)( - 4))/2( 1)

▼

Evaluate

NegNeg

- (- a)=a

x=3±sqrt((- 3)^2-4(1)(- 4))/2(1)

CalcPow

Calculate power

x=3±sqrt(9-4(1)(- 4))/2(1)

IdPropMult

Identity Property of Multiplication

x=3±sqrt(9-4(- 4))/2

MultNegNegOnePar

- a(- b)=a* b

x=3±sqrt(9+16)/2

AddTerms

Add terms

x=3±sqrt(25)/2

CalcRoot

Calculate root

x=3± 5/2

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=3± 5/2
x_1=3+5/2	x_2=3-5/2
x_1=8/2	x_2=- 2/2
x_1=4	x_2=- 1

Now, consider Equation (II). y=2x-2 We can substitute x=4 and x=- 1 into the above equation to find the values for y. Let's start with x=4.

y=2x-2

Substitute

x= 4

y=2( 4)-2

▼

Evaluate

Multiply

y=8-2

SubTerm

Subtract term

y=6

We found that y=6 when x=4. One solution of the system, which is a point of intersection of the two parabolas, is (4,6). To find the other solution, we will substitute - 1 for x in Equation (II) again.

y=2x-2

Substitute

x= - 1

y=2( - 1)-2

▼

Evaluate

MultPosNeg

a(- b)=- a * b

y=- 2-2

SubTerm

Subtract term

y=- 4

We found that y=- 4 when x=- 1. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (- 1,- 4).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (4,6). We will substitute 4 and 6 for x and y, respectively, in Equation (I) and Equation (II).

y=x^2-x-6 y=2x-2

(I), (II): x= 4, y= 6

6? = 4^2- 4-6 6? =2( 4)-2

▼

Simplify

CalcPow

(I): Calculate power

6? =16-4-6 6? =2(4)-2

Multiply

(II): Multiply

6? =16-4-6 6? =8-2

(I), (II): Add and subtract terms

6=6 ✓ 6=6 ✓

Since both equations produced true statements, the solution (4,6) is correct. Let's now check ( - 1,- 4 ).

y=x^2-x-6 y=2x-2

(I), (II): x= - 1, y= - 4

- 4? =( - 1)^2-( - 1)-6 - 4? =2( - 1)-2

▼

Simplify

CalcPow

(I): Calculate power

- 4? =1-(- 1)-6 - 4? =2(- 1)-2

SubNeg

(I): a-(- b)=a+b

- 4? =1+1-6 - 4? =2(- 1)-2

MultPosNeg

(II): a(- b)=- a * b

- 4? =1+1-6 - 4? =- 2-2

(I), (II): Add and subtract terms

- 4=- 4 ✓ - 4=- 4 ✓

Since again both equations produce true statements, the solution (- 1, - 4) is also correct.

Hints

Check the answer

Practice exercises

Asses your skill

Checking Our Answer