Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
1. Section 1.1
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Exercise 15 Page 12

Practice makes perfect
a Examining the three first diamonds, we notice that the top square is the product of the numbers in the left and right square, while the bottom square is the sum of the numbers in the left and right square.
Here, we are given the values for x and y and we want to complete the diamond. x=2 y=5 To do this we can substitute the numbers into the above pattern.
xy & (I) x+y & (II)

(I), (II): x= 2, y= 5

2 * 5 2+ 5
â–Ľ
(I), (II): Evaluate
10 2+5
10 7
Now, we can complete our first diamond.
b Here, just like in Part A, we are given values for x and y. x=- 2 y=- 1 Using the same procedure as in Part A, we can determine the contents of the diamond.
xy & (I) x+y & (II)

(I), (II): x= - 2, y= - 1

- 2( - 1) - 2+( - 1)
â–Ľ
(I), (II): Evaluate
2 - 2+(- 1)
2 - 3
Now, we can complete our second diamond.
c In this case we are given the values of the product xy and the sum x+y. We want to find the values of x and y.
xy&=8 x+y&=6 To find these values, let's think of different pairs of numbers which multiply to give us 8. Since the product of x and y is positive, we need a pair with the same sign - two positive numbers or two negative numbers. Let's also check their sums to know which, if any of these, will match the second condition as well.

c|ccc &Product&Sum 1 and 8 &8&9& * 2 and 4 &8&6& âś“ - 1 and - 8 &8&- 9& * - 2 and - 4 &8&- 6& * As we can see the pair x=2 and y=4 works for the pattern. Therefore, this is the correct pair of numbers to complete the diamond.

Notice that the pair x=4 and y=2 also works for the pattern. In this situation we have two possibilities to complete the diamond.

d Here we are given that y= 12 and xy=4. Substituting these values into our pattern gives us the following equations.
x( 12)&=4 x+ 12&= ? From the first equation we can calculate x by isolating it on one side.
x(1/2)=4
â–Ľ
LHS * 2=RHS* 2
x(1/2)(2)=8
x(1)=8
x=8
We found the value of x. Now, we substitute it into the second equation to find the square at the bottom.
x+1/2
8+1/2
â–Ľ
Evaluate
16/2+1/2
17/2
8 12
Finally, we can complete our diamond.
e In this case we are given the values of the product xy and the sum x+y. We want to find the values of x and y.
xy&=12 x+y&=- 7 To find these values, let's think of different pairs of numbers which multiply to give us 12. Since the product of x and y is positive, we need a pair with the same sign - two positive numbers or two negative numbers. Let's also check their sums to know which, if any of these, will match the second condition as well.

c|ccc &Product&Sum 1 and 12 &12&13& * 2 and 6 &12&8& * 3 and 4 &12&7& * - 1 and - 12 &12&- 13& * - 2 and - 6 &12&- 8& * - 3 and - 4 &12&- 7& âś“ As we can see the pair x=- 3 and y=- 4 works for the pattern. Therefore, this is the correct pair of numbers to complete the diamond.

Notice that the pair x=- 4 and y=- 3 also works for the pattern. In this situation we have two possibilities to complete the diamond.