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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

8. Systems of Linear and Quadratic Equations

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Exercise 2 Page 599

To solve the equation ax^2+bx+c=0, use the Quadratic Formula.

(6,10) and (- 7,192)

Practice makes perfect

We want to solve the given system of equations using the Elimination Method. y=x^2-13x+52 & (I) y=-14x+94 & (II) By subtracting Equation(II) from Equation (I), we can eliminate the y-variable. Let's do it!

y=x^2-13x+52 & (I) y=-14x+94 & (II)

(I): Subtract (II)

y- y=x^2-13x+52-( -14x+94) y=-14x+94

▼

(I): Simplify

(I): Distribute -1

y-y=x^2-13x+52+14x-94 y=-14x+94

(I): Add and subtract terms

0=x^2+x-42 y=-14x+94

(I):Rearrange equation

x^2+x-42=0 y=-14x+94

Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2+x-42=0 ⇔ 1x^2+ 1x+( - 42)=0

We can substitute a= 1, b= 1, and c= - 42 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 1)( -42))/2( 1)

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Solve for x

1^a=1

x=- 1±sqrt(1-4(1)(-42))/2(1)

a * 1=a

x=-1±sqrt(1-4(- 42))/2

MultNegNegOnePar

- a(- b)=a* b

x=-1±sqrt(1+168)/2

Add terms

x=-1±sqrt(169)/2

Calculate root

x=-1± 13/2

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± 13/2
x_1=-1+ 13/2	x_2=-1- 13/2
x_1=12/2	x_2=- 14/2
x_1=6	x_2=-7

Now, consider Equation (II). y=-14x+94 We can substitute x=6 and x=- 7 into the above equation to find the values for y. Let's start with x=6.

y=-14x+94

x= 6

y=-14( 6)+94

(- a)b = - ab

y=-84+94

Add terms

y=10

We found that y=10 when x=6. One solution to the system, which is a point of intersection of the parabola and the line, is (6,10). To find the other solution, we will substitute - 7 for x in Equation (II) again.

y=-14x+94

x= -7

y=-14( -7)+94

MultNegNegOnePar

- a(- b)=a* b

y=98+94

Add terms

y=192

We found that y=192 when x=- 7. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (-7,192).

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 (6,10). We will substitute 6 and 10 for x and y, respectively, in Equation (I) and Equation (II).

y=x^2-13x+52 & (I) y=-14x+94 & (II)

(I), (II): x= 6, y= 10

10? = 6^2-13( 6)+52 10? =-14( 6)+94

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Simplify

(I): Calculate power

10? =36-13(6)+52 10? =-14(6)+94

(I), (II): Multiply

10? =36-78+52 10? =-84+94

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

10=10 ✓ 10=10 ✓

Since both equations produced true statements, the solution (6,10) is correct. Let's now check (-7,192).

y=x^2-13x+52 & (I) y=-14x+94 & (II)

(I), (II): x= -7, y= 10

192? =( -7)^2-13( -7)+52 192? =-14( -7)+94

▼

Simplify

(I): Calculate power

192? =49-13(-7)+52 192? =-14(-7)+94

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

192? =49+91+52 192? =98+94

(I), (II): Add terms

192=192 ✓ 192=192 ✓

Since again both equations produce true statements, the solution (- 7,192) is also correct.

Subchapter links

Lesson Check p.599

Practice and Problem-Solving Exercises p.599-600

Standardized Test Prep p.601

Mixed Review p.601