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Bunuel
If 5a – 3b = –9 and 3a + b = 31, then a + b =

A) –7

B) –3

C) 6

D) 13

E) 19

Given: 5a – 3b = –9 and 3a + b = 31

Rearranging the second equation we get b = 31 - 3a

Substituting this value of b in the first equation.
5a - 3(31 - 3a) = -9 -> 14a - 93 = -9 -> 14a = 84 -> a = 6 | b = 31 - 18 = 13

Therefore, a+b = 6+13 = 19(Option E)
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5a-3b = -9.............(1)
3a+b = 31.............(2)

Multiply (1) by 3 and (2) by 5, then

15a-9b = -27
15a+5b = 155
subtract (1) from (2); 14b = 182, b = 13,
a = 6
a+b = 19
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Afc0892
5a-3b = -9.............(1)
3a+b = 31.............(2)

Multiply (1) by 3 and (2) by 5, then

15a-9b = -27
15a+5b = 155
add both, -4b = 128, b = -32,
a = 21
a+b = -11

Bunuel sir, can you check the options?

If you add you'd get 30a -4b = 128. You should subtract.
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Bunuel
Afc0892
5a-3b = -9.............(1)
3a+b = 31.............(2)

Multiply (1) by 3 and (2) by 5, then

15a-9b = -27
15a+5b = 155
add both, -4b = 128, b = -32,
a = 21
a+b = -11

Bunuel sir, can you check the options?

If you add you'd get 30a -4b = 128. You should subtract.

oh yes. silly mistake :(
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Bunuel
If 5a − 3b = −9 and 3a + b = 31, then a + b =

A. −7
B. −3
C. 6
D. 13
E. 19

Given,

3a + b = 31

b = 31 - 3a.

again

5a - 3b = -9

5a - 3(31 - 3a) = -9

5a - 93 + 9a = -9

14a = 84.

a = 6.

we know,

3a + b = 31

3*6 + b = 31

18 + b = 31

b = 31 - 18

b = 13.

Now we have ,

a = 6

b = 13.

a + b = 6 +13 = 19.

E is the correct answer.
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Is there a shorter solution?
My solution:
1) add two equations
5a - 3b + 3a + b = 31 - 9
8a - 2b = 22
4a - b = 11 => b = 4a - 11

2) find a and b by plugging b into any of them
5a - 12a + 33 = - 9
7a = 42
a = 6 => b = 31-18 = 13

Answer: a+b = 13+6 = 19. Choice E
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Bunuel
If 5a − 3b = −9 and 3a + b = 31, then a + b =

A. −7
B. −3
C. 6
D. 13
E. 19

solve for two given eqns

5a − 3b = −9 and 3a + b = 31
we get a = 6
and b = 13
a+b=19
IMO E
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