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pq+12=3p+pr, and q-r=7, what is p?

Guest Feb 26, 2015

Best Answer 

 #1
avatar+26406 
+10

pq + 12 = 3p + pr

 

Subtract pr from both sides

pq - pr + 12 = 3p

 

Factor the p from the first two terms on the left-hand side

p(q-r) + 12 = 3p

 

Use the fact that q - r = 7

7p + 12 = 3p

 

Subtract 3p from both sides

4p + 12 = 0

 

Subtract 12 from both sides

4p = -12

 

Divide both sides by 4

p = -3

.

Alan  Feb 26, 2015
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2+0 Answers

 #1
avatar+26406 
+10
Best Answer

pq + 12 = 3p + pr

 

Subtract pr from both sides

pq - pr + 12 = 3p

 

Factor the p from the first two terms on the left-hand side

p(q-r) + 12 = 3p

 

Use the fact that q - r = 7

7p + 12 = 3p

 

Subtract 3p from both sides

4p + 12 = 0

 

Subtract 12 from both sides

4p = -12

 

Divide both sides by 4

p = -3

.

Alan  Feb 26, 2015
 #2
avatar+18829 
+5

pq+12=3p+pr, and q-r=7, what is p ?

$$pq+12=3p+pr \quad | \quad :p \\\\
q + \frac{12}{p} = 3 + r \quad | \quad r = q-7 \\\\
\not{q} + \frac{12}{p} = 3 + \not{q}-7 \\\\
\frac{12}{p} = -4\\\\
p = -\frac{12}{4}\\\\
p = - 3$$

heureka  Feb 26, 2015

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