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

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

Feb 26, 2015

#1
+27336
+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

.

Feb 26, 2015

#1
+27336
+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
#2
+20831
+5

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

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

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Feb 26, 2015