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)
y- y=x^2-13x+52-( -14x+94) y=-14x+94
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(I): Simplify
y-y=x^2-13x+52+14x-94 y=-14x+94
0=x^2+x-42 y=-14x+94
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)=0We can substitute a= 1, b= 1, and c= - 42 into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 1±sqrt(1^2-4( 1)( -42))/2( 1)
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Solve for x
x=- 1±sqrt(1-4(1)(-42))/2(1)
x=-1±sqrt(1-4(- 42))/2
x=-1±sqrt(1+168)/2
x=-1±sqrt(169)/2
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
y=-14( 6)+94
y=-84+94
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
y=-14( -7)+94
y=98+94
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
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 âś“ 10=10 âś“
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
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Simplify
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 âś“ 192=192 âś“
Since again both equations produce true statements, the solution (- 7,192) is also correct.