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+10
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If twice the present age of Pedro is decreased from Petra's age 5 years ago, the result is 4. Nine times Pedro's age 3 years ago exceeds 4 times Petra's present age by 4. WHat were their ages last year?

Nov 14, 2015

#2
+6

Their ages LAST YEAR would, of course, be 1 year less:

71 -1=70 Petra's age

35 - 1=34 Pedro's age.

Nov 15, 2015

#1
+5

If twice the present age of Pedro is decreased from Petra's age 5 years ago, the result is 4. Nine times Pedro's age 3 years ago exceeds 4 times Petra's present age by 4. WHat were their ages last year?

Let pedro's age today be P

Let Petra's age today be T

2P - (T - 5)=4

9 (P-3) - 4T=4

Solve the following system:
{5+2 P-T = 4 |     (equation 1)
9 (P-3)-4 T = 4 |     (equation 2)
Express the system in standard form:
{2 P-T = -1 |     (equation 1)
9 P-4 T = 31 |     (equation 2)
Swap equation 1 with equation 2:
{9 P-4 T = 31 |     (equation 1)
2 P-T = -1 |     (equation 2)
Subtract 2/9 × (equation 1) from equation 2:
{9 P-4 T = 31 |     (equation 1)
0 P-T/9 = (-71)/9 |     (equation 2)
Multiply equation 2 by -9:
{9 P-4 T = 31 |     (equation 1)
0 P+T = 71 |     (equation 2)
Add 4 × (equation 2) to equation 1:
{9 P+0 T = 315 |     (equation 1)
0 P+T = 71 |     (equation 2)
Divide equation 1 by 9:
{P+0 T = 35 |     (equation 1)
0 P+T = 71 |     (equation 2)
Collect results:
| {P = 35 Pedro's age today
T = 71 Petra's age today

Nov 15, 2015
#2
+6