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?
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:
Answer: |
| {P = 35 Pedro's age today
T = 71 Petra's age today
Their ages LAST YEAR would, of course, be 1 year less:
71 -1=70 Petra's age
35 - 1=34 Pedro's age.
