Question

2. solve the following simultaneously equation
(1) 3a+5b=26; a+5b=22
​ {Explane me Like a Sir and How To Solve please Help me }. And In This Seatution I Can't Aatend My Class So You Now​

Answer 1

Answer:

a =2 and b = 4

Step-by-step explanation:

use elimination method -

let us consider equations

3a + 5b = 26 ------------ (i)

a + 5b = 22 ---------------(ii)

now when you subtract equation ii form first then sign of second equation will be changed....

3a + 5b = 26 ------------ (i)

- a - 5b = -22 ---------------(ii)

then you'll get..

3a + 5b- a - 5b = 26-22

2a = 4

a = 2, now by putting a = 2 in any equation you can find b

let put in equation 1

3(2) + 5b = 26

5b = 20

b = 4.

Answer 2

Answer:

the value of a =2, value of b=4

Step-by-step explanation:

Given pair of linear equations are:

3a+5b =26......(1)

a+5b=22

=>a = 22-5b.......(2) substitute the value of a in the equation (1)

then we have, 3(22-5b)+5b=26

=>66-15b+5b=26

=>66-10b=26

=>-10b=26-66

=>-10b=-40

=>b=-40/-10

therefore, b=4

now we substitute the value of b in the equation (2),then we get

a=22-5(4)

=>a=22-20

therefore a=2

check:

(3×2)+(5×4)=6+20=26

2+(5×4)=2+20=22