Question

Solve by substitution method 2 x+3 y-2=0 and 5 x-3 y/2 -2=0

Answer 1

x = (1/2) = 0.50

y = (1/3) = 0.33

Step-by-step explanation:

We are given the following pair of equations :- $2x+3y-2=0$  and  $5x-\frac{3y}{2} -2=0$ .

And we have to find the value of x and y using substitution method.

Substitution method means that taking the value of one variable from the equation and then substituting that value of variable in another equation.

First equation is given by;

                            $2x+3y-2=0$

                            $2x=2 - 3y$

                              $x=\frac{2-3y}{2}$  ------------------ [Equation 1]

Second equation is given by;

                             $5x-\frac{3y}{2} -2=0$  ----------------- [Equation 2]

Now, substituting the value of x from equation 1 into equation 2, we get;

                             $5(\frac{2-3y}{2}) -\frac{3y}{2} -2=0$    

                              $\frac{10-15y}{2} -\frac{3y}{2} =2$

                               $\frac{10-15y-3y}{2} =2$

                                $\frac{10-18y}{2} =2$

                               $10-18y =2 \times 2$      

                               $10-18y =4$

                                  $18y =10-4$

                                  $18y =6$

                                   $y=\frac{6}{18} = \frac{1}{3}$

Now, putting value of y in equation 1, we get;    

                                  $x=\frac{2-3y}{2}$

                                  $x=\frac{2-3\times (\frac{1}{3}) }{2}$

                                  $x=\frac{2-1 }{2}$ =  $\frac{1}{2}$

                                     

Therefore, value of x =  $\frac{1}{2}$  = 0.50 and  y =  $\frac{1}{3}$  = 0.33.

Answer 2

Step-by-step explanation:

x = (1/2) = 0.50

y = (1/3) = 0.33

Step-by-step explanation:

We are given the following pair of equations :- 2x+3y-2=02x+3y−2=0 and 5x-\frac{3y}{2} -2=05x−

2

3y

−2=0 .

And we have to find the value of x and y using substitution method.

Substitution method means that taking the value of one variable from the equation and then substituting that value of variable in another equation.

First equation is given by;

2x+3y-2=02x+3y−2=0

2x=2 - 3y2x=2−3y

x=\frac{2-3y}{2}x=

2

2−3y

------------------ [Equation 1]

Second equation is given by;

5x-\frac{3y}{2} -2=05x−

2

3y

−2=0 ----------------- [Equation 2]

Now, substituting the value of x from equation 1 into equation 2, we get;

5(\frac{2-3y}{2}) -\frac{3y}{2} -2=05(

2

2−3y

)−

2

3y

−2=0

\frac{10-15y}{2} -\frac{3y}{2} =2

2

10−15y

−

2

3y

=2

\frac{10-15y-3y}{2} =2

2

10−15y−3y

=2

\frac{10-18y}{2} =2

2

10−18y

=2

10-18y =2 \times 210−18y=2×2

10-18y =410−18y=4

18y =10-418y=10−4

18y =618y=6

y=\frac{6}{18} = \frac{1}{3}y=

18

6

=

3

1

Now, putting value of y in equation 1, we get;

x=\frac{2-3y}{2}x=

2

2−3y

x=\frac{2-3\times (\frac{1}{3}) }{2}x=

2

2−3×(

3

1

)

x=\frac{2-1 }{2}x=

2

2−1

= \frac{1}{2}

2

1

Therefore, value of x = \frac{1}{2}

2

1

= 0.50 and y = \frac{1}{3}

3

1

= 0.33.