Question

solve the following simultaneous equations.. 1. 3a+5b=26;a+5b=22l​

Answer 1

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3a+5b=26…………(1)

a+5b=22……………(2)

Now subtract equⁿ(1)&(2)

3a+5b=26

- a+5b=22

2a = 4

a=$\frac{4}{2}$

a=2

Therefore a=2

Now, put a=2 in equⁿ(2)

a+5b=22

2+5b=22

5b=22-2

5b=20

b=$\frac{20}{5}$

b=4

Therefore (a, b) is (2,4)

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Answer 2

3a + 5b = 26; .....(I)

a + 5b = 22 .....(II)

Subtracting (II) from (I)

2a = 4

⇒ a = 2

Putting the value of a = 2 in (II)

5b = 22 - 2 = 20

⇒b = 20/5

=4

Thus, a = 2 and b = 4.