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Core Connections Algebra 2, 2013 View details

1. Section 11.1

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Exercise 22 Page 575

Practice makes perfect

a To solve the given system of equations we can use substitution. When using this method, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
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 the x-variable in the first equation.

x-2y=7 & (I) 2x+y=3 & (II)

AddEqn

(I): LHS+2y=RHS+2y

x=7+2y 2x+y=3

Now that we have isolated x, we can solve the system by substitution.

x=7+2y 2x+y=3

Substitute

(II):x= 7+2y

x=7+2y 2( 7+2y)+y=3

Distr

(II):Distribute 2

x=7+2y 14+4y+y=3

SubEqn

(II):LHS-14=RHS-14

x=7+2y 4y+y=-11

AddTerms

(II):Add terms

x=7+2y 5y=-11

DivEqn

(II): .LHS /5.=.RHS /5.

x=7+2y y= -115

MoveNegNumToFrac

(II):Put minus sign in front of fraction

x=7+2y y=- 115

Great! Now, to find the value of x we need to substitute y=- 115 into either one of the equations in the given system. Let's use the first equation.

x=7+2y y=- 115

Substitute

(I):y= - 115

x=7+2( - 115) y=- 115

MultPosNeg

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

x=7-2* 115 y=- 115

MoveLeftFacToNum

(I): a*b/c= a* b/c

x=7- 225 y=- 115

NumberToFrac

(I): a = 5* a/5

x= 355- 225 y=- 115

SubFrac

(I):Subtract fractions

x= 135 y=- 115

The solution, or point of intersection, to this system of equations is the point ( 135,- 115).

b We want to solve the given system of equations.

x+4y 3- 6y-x 4=-3 & (I) x 10+5y=2 & (II) We will solve the system of equations by multiplying each side of the equations by the least common denominator to clear the denominators.

x+4y 3- 6y-x 4=-3 x10+5y=2

MultEqn

(I): LHS * ( 3* 4)=RHS* ( 3* 4)

( x+4y 3- 6y-x 4)*( 3* 4)=-3* ( 3* 4) x10+5y=2

▼

Simplify

Multiply

(I):Multiply

(x+4y)*(3*4)3- (6y-x)*(3*4)4=-36 x10+5y=2

CancelCommonFac

(I):Cancel out common factors

(x+4y)*(3*4)3- (6y-x)*(3*4)4=-36 x10+5y=2

SimpQuot

(I): Simplify quotient

(x+4y)* 4 - (6y-x)*3 = -36 x10+5y=2

Distr

(I): Distribute 4

4x+16y-(6y-x)*3 = -36 x10+5y=2

CommutativePropMult

(I): Commutative Property of Multiplication

4x+16y-3(6y-x) = -36 x10+5y=2

Distr

(I): Distribute -3

4x+16y-18y+3x = -36 x10+5y=2

AddSubTerms

(I):Add and subtract terms

7x-2y=-36 x 10+5y=2

MultEqn

(II):LHS * 10=RHS* 10

7x-2y=-36 ( x 10+5y)*10=2*10

Multiply

(II):Multiply

7x-2y=-36 x10*10+50y=20

FracMultDenomToNumber

(II):a/10* 10 = a

7x-2y=-36 x+50y=20

Now we are prepared to solve the system of equations. To do this, let's use the Substitution Method again! Observing the given equations, it looks like it will be simplest to isolate the x-variable in the second equation.

7x-2y=-36 x+50y=20

SubEqn

(II): LHS-50y=RHS-50y

7x-2y=-36 x=20-50y

Now that we have isolated x, we can solve the system by substitution.

7x-2y=-36 x=20-50y

Substitute

(I):x= 20-50y

7( 20-50y)-2y=-36 x=20-50y

Distr

(I):Distribute 7

140-350y-2y=-36 x=20-50y