Question

solve the following system of linear equation by substitution method:. 2x-y=2. x+3y=15​

Answer 1

Answer:

2x-y=2

x+3y=15

multiply eq 1 by 3 we get

6x-3y=6

x+3y=15

=7x=21

x=3

put in eq 1 we get

6-y=2

-y=2-6

y=4

Answer 2

Given :-

• 2x - y = 2 ⇒ Equation 1
• x + 3y = 15 ⇒ Equation 2

Aim :-

• To find the values of x and y

Answer :-

Equation 1 :-

⇒ 2x - y = 2

Transposing (-y) to the other side,

⇒ 2x = 2 + y

Transposing 2,

$\implies \sf x = \dfrac{2 + y}{2}$

Equation 2 :-

Substituting the value of x as derived from the above equation,

$\implies\sf \dfrac{2 + y}{2} + 3y = 15$

Taking LCM = 2

$\implies \sf \dfrac{2 + y + 6y}{2} = 15$

Adding the numbers in the numerator,

$\implies \sf \dfrac{2 + 7y}{2} = 15$

Transposing 2 to the other side,

⇒ 2 + 7y = 15 × 2

⇒ 2 + 7y = 30

Transposing 2,

⇒ 7y = 30 - 2

⇒ 7y = 28

Transposing 7,

$\implies \sf y = \dfrac{28}{7}$

Reducing to the lowest terms,

⇒ y = 4

Now that we have the value of y, x :-

$\implies \sf x = \dfrac{2 + 4}{2}$

$\implies \sf x = \dfrac{6}{2}$

Reducing to the lowest terms,

⇒ x = 3

Therefore x = 3 and y = 4

Verification :-

Let us verify the answer by substituting the values of x and y as 3 and 4 respectively in the 2 equations.

Equation 1 :-

⇒ 2(3) - 4 = 2

LHS :-

⇒ 2(3) - 4

⇒ 6 - 4

⇒ 2

RHS :-

⇒ 2

LHS = RHS

Equation 2 :-

⇒ 3 + 3(4) = 15

LHS :-

⇒ 3 + 3(4)

⇒ 3 + 12

⇒ 15

RHS :-

⇒ 15

LHS = RHS

Hence verified