Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
3. Section 2.3
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Exercise 83 Page 80

Practice makes perfect
a Comparing this diamond to the one showing the pattern, we can create two equations.
y=& -1 x+y=& 10 Since we know that y=-1, we can substitute it in the second equation to find x.
x+y=10
x+( -1)=10
x-1=10
x=11
Since we know that x=11, we can find the upper square of the diamond by substituting x=11 and y=-1 in the expression xy.
xy
11( -1)
-11
Now we can complete our diamond!
b In this diamond we have been given the values of x and y.
&x= 5 &y= -3

Let's determine xy and x+y by substituting these values into the expressions. xy:& 5( -3)=-15 x+y:& 5+( -3)=2 Now we can complete our diamond!

c With the given information we can create the following equations.
x=& 23 xy=& 1 To complete the diamond we should substitute x= 23 in the second equation and find y.
xy=1
2/3* y =1
Simplify
2y/3=1
2y=3
y=3/2
Having found y, we can determine the last square in the diamond by substituting x and y in x+y.
x+y
2/3+ 3/2
Simplify
4/6+3/2
4/6+9/6
13/6
Finally, we can complete our diamond!
d For the last diamond we can create the following equations.
xy=& -14 x+y=& -5 This tells us that x and y, when multiplied, must give us a negative number. Therefore, either x or y must be negative and the other one must be positive. We have two cases that both results in a product of -14 when multiplied.

-14=-2( 7) -14= 2(-7) To determine which pair of integers is correct, we have to check which one sums to-5. -2 +7 &= 5 & * 2+(-7)&=-5 & ✓ As we can see, the values of x and y are 2 and -7. Now we can complete our diamond!