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Core Connections Algebra 1, 2013 View details

2. Section 1.2

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Exercise 33 Page 19

Use the given pattern to complete each of the diamonds.

Practice makes perfect

To find the missing numbers, let's identify the variables using our variable diamond and substitute the correct numbers into our pattern.

Diamond A

In Diamond A, we are given that C=-98 and A=7. Substituting these values into our pattern gives us the following equations. -98 &=7 * B D&= 7+B From the first equation, we can calculate B by isolating it on one side.

-98=7* B

DivEqn

.LHS /7.=.RHS /7.

-14=B

RearrangeEqn

Rearrange equation

A=-14

We found B. Now, we can substitute it into the second equation to find D.

D=7+B

Substitute

B= -14

D=7+( -14)

AddNeg

a+(- b)=a-b

D=7-14

SubTerm

Subtract term

D=-7

Now we can complete our first diamond.

Diamond B

In Diamond B, we are given values for A and B, and we want to find C and D. A= 3/2 and B=3/2 To do this, we can substitute the numbers in our pattern.

C= A * B D=A+B

(I), (II): A= 3/2, B= 3/2

C= 32 * 32 D= 32+ 32

MultFrac

(I): Multiply fractions

C= 3*32*2 D= 32+ 32

Multiply

(I):Multiply

C= 94 D= 32+ 32

▼

(I):Write fraction as a mixed number

Rewrite

(I):Rewrite 9 as 8+1

C= 8+14 D= 32+ 32

WriteSumFrac

(I):Write as a sum of fractions

C= 84+ 14 D= 32+ 32

CalcQuot

(I):Calculate quotient

C=2+ 14 D= 32+ 32

AddTerms

(I):Add terms

C=2 14 D= 32+ 32

AddFrac

(II): Add fractions

C=2 14 D= 62

CalcQuot

(II):Calculate quotient

C=2 14 D=3

We can now complete our second diamond.

Diamond C

In this case, we are given that C= 54 and B=- 52. Substituting these values into our pattern gives us the following equations. 5/4&=A * (-5/2) D &= A+ (-5/2) From the first equation, we can calculate A by isolating it on one side.

5/4=A* (-5/2)

MultPosNeg

a(- b)=- a * b

5/4=-5/2A

▼

Isolate A

MultEqn

LHS * 2=RHS* 2

5/4*2=-5A

DivEqn

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

5/4*2÷(-5)=A

MoveRightFacToNum

a/c* b = a* b/c

5*2/4÷(-5)=A

DivFracSL

.a/b /c.= a/b* c

5*2/4*(-5)=A

Multiply

10/-20=A

ReduceFrac

a/b=.a /10./.b /10.

1/-2=A

MoveNegDenomToFrac

Put minus sign in front of fraction

-1/2=A

RearrangeEqn

Rearrange equation

A=-1/2

We found A. Now, we can substitute it into the first equation to find D.

D=A+(-5/2)

Substitute

A= -1/2

D= -1/2+(-5/2)

AddNeg

a+(- b)=a-b

D=-1/2-5/2

SubFrac

Subtract fractions

D=-1-5/2

SubTerm

Subtract term

D=-6/2

CalcQuot

Calculate quotient

D=-3

Now we can complete our third diamond.

Diamond D

In Diamond D, we are given that D=0 and A=-10. Substituting these values into our pattern gives us the following equations. C &=-10 * B 0&= -10 +B From the second equation, we can calculate B by isolating it on one side.

0=-10+B

AddEqn

LHS+10=RHS+10

10=B

RearrangeEqn

Rearrange equation

B=10

We found B. Now, we can substitute it into the second equation to find D.

C=-10* B

Substitute

B= 10

C=-10* 10