Exercise 8 Page 149

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 values for A and B, and we want to find C and D. A= 9 and B=-9 To do this, we can substitute the numbers in our pattern.

C= A * B D=A+B

(I), (II): A= 9, B= -9

C= 9 * ( -9) D= 9+( -9)

Multiply

(I): Multiply

C=-81 D=9+(-9)

AddNeg

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

C=-81 D=0

Now we can complete our first diamond.

Diamond B

Here we are given C= -36 and D=0. We want to find the numbers A and B. -36=A* B 0=A+B Let's think of the possible numbers for A and B that would give us -36 when we multiply them. &-36=-1* 36 &-36=-2* 18 &-36=-3 * 12 &-36=-4 * 9 &-36=-6 * 6 Let's check which, if any of these, will match the second condition as well. &-1+ 36 =35 * &-2 + 18=16 * &-3+ 12 =9 * &-4 + 9=5 * & -6+6 =0 The pair A=-6 and B=6 works for the pattern. Therefore, this is the correct pair of numbers to complete the diamond.

Diamond C

In this case, we are given that C=-20 and A=4. Substituting these values into our pattern gives us the following equations. -20&=4 * B D&= 4+B From the first equation, we can calculate B by isolating it on one side.

-20=4 * B

DivEqn

.LHS /4.=.RHS /4.

-5=B

RearrangeEqn

Rearrange equation

B=-5

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

D=4+B

Substitute

B= -5

D=4+( -5)

AddNeg

a+(- b)=a-b

D=4-5

SubTerm

Subtract term

D=-1

Now we can complete our third diamond.

Diamond D

In this case, we are given that D= 12 and B= 45. Substituting these values into our pattern gives us the following equations. C&=A * 45 12&= A+ 45 From the second equation, we can calculate A by isolating it on one side.

1/2=A +4/5

SubEqn

LHS-4/5=RHS-4/5

1/2-4/5=A

ExpandFrac

a/b=a * 5/b * 5

5/10-4/5=A

ExpandFrac

a/b=a * 2/b * 2

5/10-8/10=A

SubFrac

Subtract fractions