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Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 4.1
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Exercise 40 Page 180

Practice makes perfect
a In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using this method, there are three steps.
  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.
Let's use (II) for our initial substitution.
y=3x-5 & (I) y=-2x-15 & (II)
Substitute

(I):y= -2x-15

-2x-15=3x-5 y=-2x-15
â–Ľ
(I):Solve for x
SubEqn

(I):LHS-3x=RHS-3x

-5x-15=-5 y=-2x-15
AddEqn

(I):LHS+15=RHS+15

-5x=10 y=-2x-15
DivEqn

(I): .LHS /(-5).=.RHS /(-5).

x=-2 y=-2x-15
Great! Now, to find the value of y, we need to substitute x=-2 into either one of the equations in the given system. Let's use the second equation.
x=-2 y=-2x-15
Substitute

(II):x= -2

x=-2 y=-2( -2)-15
MultNegNegOnePar

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

x=-2 y=4-15
SubTerm

(II): Subtract term

x=-2 y=-11
The solution to this system of equations is the point (-2,-11). That means the graphs of given equations intersect at exactly one point.
b In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using this method, there are three steps.
  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.
y-7=-2x & (I) 4x+2y=14 & (II)
AddEqn

(I):LHS+7=RHS+7

y=7-2x 4x+2y=14
Now that we've isolated y, we can solve the system by substitution.
y=7-2x 4x+2y=14
Substitute

(II):y= 7-2x

y=7-2x 4x+2( 7-2x)=14
Distr

(II):Distribute 2

y=7-2x 4x+14-4x=14
SubTerm

(II):Subtract term

y=7-2x 14=14
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.
c We want to solve the given system of equations using the Substitution Method.
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.
y=2(x+3)^2-5 y=14x+17
Substitute

(I): y= 14x+17

14x+17=2(x+3)^2-5 y=14x+17
â–Ľ
(I): Simplify
ExpandPosPerfectSquare

(I): (a+b)^2=a^2+2ab+b^2

14x+17=2(x^2+6x+9)-5 y=14x+17
Distr

(I):Distribute 2

14x+17=2x^2+12x+18-5 y=14x+17
SubTerm

(I): Subtract term

14x+17=2x^2+12x+13 y=14x+17
SubEqn

(I): LHS-14x=RHS-14x

17=2x^2-2x+13 y=14x+17
SubEqn

(I): LHS-17=RHS-17

0=2x^2-2x-4 y=14x+17
DivEqn

(I):.LHS /2.=.RHS /2.

0=x^2-x-2 y=14x+17
RearrangeEqn

(I):Rearrange equation

x^2-x-2=0 y=14x+17
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)
â–Ľ
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
â–Ľ
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
â–Ľ
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.
d We want to solve the given system of equations using the Substitution Method.
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.
y=3(x-2)^2+3 y=6x-12
Substitute

(I): y= 6x-12

6x-12=3(x-2)^2+3 y=6x-12
â–Ľ
(I): Simplify
ExpandNegPerfectSquare

(I): (a-b)^2=a^2-2ab+b^2

6x-12=3(x^2-4x+4)+3 y=6x-12
Distr

(I):Distribute 3

6x-12=3x^2-12x+12+3 y=6x-12
AddTerms

(I): Add terms

6x-12=3x^2-12x+15 y=6x-12
SubEqn

(I): LHS-6x=RHS-6x

-12=3x^2-18x+15 y=6x-12
AddEqn

(I): LHS+12=RHS+12

0=3x^2-18x+27 y=6x-12
DivEqn

(I):.LHS /3.=.RHS /3.

0=x^2-6x+9 y=6x-12
RearrangeEqn

(I):Rearrange equation

x^2-6x+9=0 y=6x-12
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)
â–Ľ
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
â–Ľ
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.
Subchapter links expand_more
arrow_right 4.1.1 Strategies for Solving Equations p.171-172
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Strategies for Solving Equations 10 (Page 171)
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Strategies for Solving Equations 12 (Page 172)
Strategies for Solving Equations 13 (Page 172)
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arrow_right 4.1.2 Solving Equations and Systems Graphically p.176-178
Solving Equations and Systems Graphically 22 (Page 176)
Solving Equations and Systems Graphically 23 (Page 176)
Solving Equations and Systems Graphically 24 (Page 176)
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Solving Equations and Systems Graphically 27 (Page 176)
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Solving Equations and Systems Graphically 29 (Page 176)
Solving Equations and Systems Graphically 30 (Page 177)
Solving Equations and Systems Graphically 31 (Page 177)
Solving Equations and Systems Graphically 32 (Page 177)
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Solving Equations and Systems Graphically 34 (Page 177)
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arrow_right 4.1.3 Finding Multiple Solutions to Systems of Equations p.180-181
Finding Multiple Solutions to Systems of Equations 40 (Page 180)
Finding Multiple Solutions to Systems of Equations 41 (Page 180)
Finding Multiple Solutions to Systems of Equations 42 (Page 180)
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Finding Multiple Solutions to Systems of Equations 44 (Page 181)
Finding Multiple Solutions to Systems of Equations 45 (Page 181)
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arrow_right 4.1.4 Using Systems of Equations to Solve Problems p.184
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Using Systems of Equations to Solve Problems 53 (Page 184)
Using Systems of Equations to Solve Problems 54 (Page 184)
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