Question

How do you solve the simultaneous equations 7a−3b=17 and 2a+b=16?

Answer 1

$\huge \mathfrak{hello \: there}$

7a-3b=17-----(1)

2a+b=16------(2)

Multiplying equation (1) by 2 and eq.(2) by 7,

➡14a-6b=34-----(3)

➡14a+7b=112-----(4)

Subtracting (4) from (3),

-13b=-78

b=6

Substituting the above value of b in (1),

7a-3(6)=17

7a-18=17

7a=35

a=5

Thus, a=5, b=6

Answer 2

7a-3b=17

so first find out a interms of b

... 7a=17+3b

=> a= (17+3b)/7...(1)

Now put this value of a in the second equation.....

2a+b=16

=> 2(17+3b)/7 + b = 16

=> (34+6b)/7+b = 16....take the LCM of 7

=> (34+6b+7b)/7= 16

=> 34+13b = 16 *7

=> 13b = 112 -34

=> b = 78/13

= b = 6

So you got B now find out a from eq (1)

or any of the given eq

let's take-> 2a+b=16

=>2a+6=16

=>2a= 10

=>a = 5

Hence you got your answer

a as 5 and b as 6....

Hope my answer helps....