Question

11. Solve 2a + 3b = 18 and 3a + 2b =17​

Answer 1

$\text{Given equations are}$

$2a+3b=18$........(1)

$3a+2b=17$.........(2)​

$\text{Addding (1) and (2), we get }$

$5a+5b=35$

$a+b=7$.........(3)​

$\text{subtracting (2) from (1), we get }$

$-a+b=1$.........(4)​

$\text{Addding (3) and (4), we get }$

$2b=8$

$\implies\bf\;b=4$

$\text{Put b=4 in (3), we get}$

$a+4=7$

​

$\implies\bf\;a=3$

$\therefore\text{The solution is a=3 and b=4}$

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Answer 2

Step-by-step explanation:

We have,

2a + 3b = 18 ....(1)

and

3a + 2b =17​ ...(2)

Multiplying equation (1) be 3 and equation (2) by 2, such that,

6a + 9b = 54 ....(3)

6a + 4b = 34 ...(4)

Now subtracting equation (3) and (4) such that,

9b - 4b = 54 - 34

5b = 20

b = 4

Put b = 4 in equation (1) such that,

2a + 3(4) = 18

2a = 18 - 12

2a = 6

a = 3

So, the value of a is 3 and that of b is 4.

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