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klasskru [66]
3 years ago
7

Find the value of p3- q3, if p-q = -7 and pq= -10

Mathematics
1 answer:
sergiy2304 [10]3 years ago
5 0
<h3>Answer:   -133</h3>

===================================================

Work Shown:

p-q = -7

(p-q)^2 = (-7)^2

p^2-2pq+q^2 = 49

p^2-2(-10)+q^2 = 49

p^2+20+q^2 = 49

p^2+q^2 = 49-20

p^2+q^2 = 29

By the difference of cubes formula, we can say,

p^3 - q^3 = (p-q)*(p^2+pq+q^2)

p^3 - q^3 = (p-q)*(p^2+q^2+pq)

p^3 - q^3 = (-7)*(29-10)

p^3 - q^3 = -133

--------------

Here's another way to solve:

p-q = -7

p = q-7

pq = -10

(q-7)*q = -10

q^2-7q = -10

q^2-7q+10 = 0

(q-5)(q-2) = 0

q-5 = 0 or q-2 = 0

q = 5 or q = 2

If q = 5, then p = q-7 = 5-7 = -2

If q = 2, then p = q-7 = 2-7 = -5

We can see that no matter what we pick for q, we'll get p*q = -10

Also, you should find that p^3-q^3 is equal to -133 for either case as well.

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For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x
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The <u>correct answer</u> is:

D) \left \{ {{2x-y=7} \atop {2x+7y=31}} \right..

Explanation:

We solve each system to find the correct answer.

<u>For A:</u>
\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.

Since we have the coefficients of both variables the same, we will use <u>elimination </u>to solve this.  

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. &#10;\\&#10;\\6x=23

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is <u>not the x-coordinate</u> of the answer we are looking for, so <u>A is not correct</u>.

<u>For B</u>:
\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.

For this equation, it will be easier to isolate a variable and use <u>substitution</u>, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>B is not correct</u>.

<u>For C</u>:
Since the coefficient of x in the second equation is 1, we will use <u>substitution</u> again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
3y=-35

Divide both sides by 3:
3y/3 = -35/3
y = -35/3

This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>C is not correct</u>.  

<u>For D</u>:
Since the coefficients of x are the same in each equation, we will use <u>elimination</u>.  We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. &#10;\\&#10;\\-8y=-24

Divide both sides by -8:
-8y/-8 = -24/-8
y=3

The y-coordinate is correct; next we check the x-coordinate  Substitute the value for y into the first equation:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so <u>D is the correct answer</u>.
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