Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 1.3
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Exercise 112 Page 59

Practice makes perfect
a In this diamond we have been given the values of x and y.

&x= 8 &y= -8Let's determine xy and x+y by substituting these values into the expressions. xy:& 8( -8)=-64 x+y:& 8+( -8)=0 Now we can complete our diamond!

b For this diamond we can create the following two equations.

xy=& -25 x+y=& 0This 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 only one case that results in a product of -25 when multiplied. -25=-5(5) As we can see, the values of x and y are 5 and -5. Now we can complete our diamond!

c With the given information we can create the following equations.

x=& 4 xy=& -12 To complete the diamond we should substitute x=4 in the second equation and find y.

xy=-12
4* y =-12
y=-3

Having found y, we can determine the last square in the diamond by substituting x and y in x+y.

x+y
4+( -3)
4-3
1

Finally, we can complete our diamond!

d In this diamond we have been given the values of x and y.

&x= 1/2, y= -1/3Let's determine xy and x+y by substituting these values into the expressions. xy:& 1/2( -1/3)=-1/6 [1em] x+y:& 1/2+( -1/3)=1/6 Now we can complete our diamond!