Question

Plz solve the following equation by substitution method​

Answer 1

Given :

• $\sf \dfrac {x}{2}+\dfrac{2y}{3}=-1$
• $\sf x - \dfrac{y}{3}=3$

To Find :

• Solve the equations by using substition method.

Solution :

Simplifying the equations,

$\begin{array}{c|c}\sf \dfrac {x}{2}+\dfrac{2y}{3}=-1& \sf x - \dfrac{y}{3}=3 \\\\\sf \dfrac{3x+4y}{6} =-1&\sf\dfrac{3x-y}{3}=3\\\\\sf 3x+4y=-6 ...(1)&\sf 3x-y=9 ...(2)\end{array}$

Calculating value of one of the variables,

$\Rightarrow \sf 3x-y=9$

$\Rightarrow \sf -y=9-3x$

$\Rightarrow \sf y = 3x-9 ...(3)$

Substituting eqn(3) in eqn (1),

$\longrightarrow \sf 3x+4(3x-9)=-6\\\\\sf \longrightarrow 3x+12x-36=-6\\\\\longrightarrow \sf 15x=-6+36\\\\\longrightarrow \sf 15x=30\\\\\longrightarrow \sf x = \dfrac{30}{15}\\\\\longrightarrow \sf \boxed{x=2}$

Substitute the value of x in (2),

$\longrightarrow \sf 3x-y=9\\\\\longrightarrow \sf 3(2)-y=9\\\\\longrightarrow \sf 6-y=9\\\\\longrightarrow \sf -y=9-6\\\\\longrightarrow \sf -y=3\\\\\longrightarrow \sf \boxed{y=-3}$

Verifying the values in eqn(1)&(2),

${\begin{array}{c|c}\sf 3x+4y=-6&\sf 8x-y=9\\\\\sf3(2)+4(-3)=-6&\sf3(2)-(-3)=9\\\\\sf6-12=-6&\sf6+3=9\\\\\sf-6=-6&\sf9=9\end{array}}\\\\ \sf Hence\: Verified$

Required Answer :

• The value of x is $\underline{\sf 2}$
• The value of y is $\underline{\sf -3}$

Answer 2

Step-by-step explanation:

x/2 + 2y/3 = -1 x - y/3 = 3

• SIMPLiFy BoTH EQUATION

3x + 4y = −6 _(i) 3x − y = 9 _(ii)

• SUBTRACT → (i) - (ii) , we get ,

5y = -15

y = -3

• PUTTiNG y = -3 in (i)

3x + 4(-3) = −6

3x + (-12) = −6

3x = −6 + 12

x = 2