Question

Algebra

5p+7p=26 ; 2p+3q=11. Solve by elimination method.​

Answer 1

Answer:

5p+7q=26 equation 1

2p+3q=11 equation 2

Multiply equation 1 by 2 and equation 2 by 5 then subtract 2nd equation from 1st, you will get

14q-15q=52-55

-q=-3

q=3

put this value of q in any of the above equations

you will get p=11

Answer 2

★ Correct Question :-

5p+7q=26

2p+3q=11

Solve by elimination method.

★ Solution :-

$\sf \leadsto 5p + 7q = 26 \: \: --- (i)$

$\sf \leadsto 2p + 3q = 11 \: \: --- (ii)$

By multiplying the first equation by 3 and the second by 7.

$\sf \leadsto 3(5p + 7q = 26)$

$\sf \leadsto 15p + 21q = 78$

$\sf \leadsto 7(2p + 3q = 11)$

$\sf \leadsto 14p + 21q = 77$

Now, by subtracting the first equation by second.

$\sf \: \: \: 15p + 21q = 78 \\ \sf −14p + 21q = 77$

$\sf \leadsto 1p - 0 = 1$

$\sf \leadsto 1p = 1$

$\sf \leadsto p = 1$

Now, by substituting the value of p in first equation.

$\sf \leadsto 5p + 7q = 26$

$\sf \leadsto 5(1) + 7q = 26$

$\sf \leadsto 5 + 7q = 26$

$\sf \leadsto 7q = 26 - 5$

$\sf \leadsto 7q = 21$

$\sf \leadsto q = \dfrac{21}{7}$

$\sf \leadsto q = 3$

Therefore, the value of p and q are 1 and 3 respectively.