Exercise 115 Page &nbsp; 327

Distribute 15

15(1/5x)+15(1/3x)=15(2)

MovePartNumRight

a* b/c=a/c* b

15(1)/5x+15(1)/3x=15(2)

15/5x+15/3x=30

Calculate quotient

3x+5x=30

AddTerms

Add terms

8x=30

Now we can use the Division Property of Equality to isolate x.

8x=30

DivEqn

.LHS /8.=.RHS /8.

8x/8=30/8

Calculate quotient

x=3.75

To solve the given equation we will use the Multiplication Property of Equality to get rid of the decimals.

x+0.15x=$ 2

MultEqn

LHS * 100=RHS* 100

100(x+0.15x)=100($ 2)

Distribute 100

100x+100(0.15x)=100($ 2)

100x+15x=$ 200

AddTerms

Add terms

115x=$200

Now we can use the Division Property of Equality to isolate x.

115x=$200

DivEqn

.LHS /115.=.RHS /115.

115x/115=$200/115

Calculate quotient

x=$ 1.739130...

RoundDec

Round to 2 decimal place(s)

x≈ $ 1.74

We are asked to solve the following equation. x+2/3=x-2/7Let's first get rid of the fractions by using cross multiplication.

x+2/3=x-2/7

CrossMult

Cross multiply

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

Distribute 3

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

Distribute 7

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

3x-6=7x+14

Now we will group all the x-terms and constant terms together on opposite sides of the equation using properties of equality.

3x-6=7x+14

SubEqn

LHS-3x=RHS-3x

-6 = 4x+14

SubEqn

LHS-14=RHS-14

- 20 = 4x

DivEqn

.LHS /4.=.RHS /4.

-20/4=4/4

Calculate quotient

- 5=x

RearrangeEqn

Rearrange equation

x=- 5

We are asked to solve the following system of equations. y= 23x+8 & (I) y= 12x+10 & (II) Since both equations are solved for y, we will use the Substitution Method to solve the system. Let's substitute Equation (I) for y into Equation (II).

y= 23x+8 & (I) y= 12x+10 & (II)

Substitute

(II): y= 23x+8

y= 23x+8 23x+8= 12x+10

MultEqn

(II): LHS * 6=RHS* 6

y= 23x+8 6( 23x+8)=6( 12x+10)

(II): Distribute 6

y= 23x+8 6( 23x)+6(8)=6( 12x)+6(10)