Question

solve graphically 3x+2y=1 2x-y =3​

Answer 1

Given

3x + 2y = 1 (i)

2x - y = 3 (ii)

Now Take

3x + 2y = 1 (i)

When x = 0

3(0) + 2y = 1

2y = 1

y = 1/2

We get point ( 0 , 1/2)

When y = 0

3x + 2(0) = 1

3x = 1

x = 1/3

We get point (1/3 , 0)

Now Take

2x - y = 3

When x = 0

2(0) - y = 3

y = - 3

We get point (0,-3)

When y = 0

2x - 0 = 3

x = 3/2

We get point ( 3/2 , 0)

Answer 2

Consider Line (1),

$\rm :\implies\: \boxed{ \pink{ \bf \: 3x + 2y\: = \tt \: 1}}$

❶ Substituting 'x = 0' in the given equation, we get

$\rm :\implies\:3 \times 0 + 2y = 1$

$\rm :\implies\:0 + 2y = 1$

$\rm :\implies\:y = \dfrac{1}{2}$

$\rm :\implies\:y \: = \: 0.5$

❷ Substituting 'x = - 1' in the given equation, we get

$\rm :\implies\:3 \times ( - 1) + 2y = 1$

$\rm :\implies\: - 3 + 2y = 1$

$\rm :\implies\:2y = 1 + 3$

$\rm :\implies\:2y = 4$

$\rm :\implies\:y = 2$

❸ Substituting 'x = - 2' in the given equation, we get

$\rm :\implies\:3 \times ( - 2) + 2y = 1$

$\rm :\implies\: - 6 + 2y = 1$

$\rm :\implies\:2y = 7$

$\rm :\implies\:y = \dfrac{7}{2}$

$\rm :\implies\:y = 3.5$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0.5 \\ \\ \sf - 1 & \sf 2 \\ \\ \sf - 2 & \sf 3.5 \end{array}} \\ \end{gathered}$

➢ See the attachment, Red line represents 3x + 2y = 1

Now,

Consider Line (2),

$\rm :\implies\: \boxed{ \pink{ \bf \: 2x \: - \: y \: = \tt \: 3}}$

❶ Substituting 'x = 0' in the given equation, we get

$\rm :\implies\:2 \times (0) - y = 3$

$\rm :\implies\:0 - y = 3$

$\rm :\implies\:y = - 3$

❷ Substituting 'x = 1' in the given equation, we get

$\rm :\implies\:2 \times 1 - y = 3$

$\rm :\implies\:2 - y = 3$

$\rm :\implies\:y = - \: 1$

❸ Substituting 'x = 2' in the given equation, we get

$\rm :\implies\:2 \times 2 - y = 3$

$\rm :\implies\:4 - y = 3$

$\rm :\implies\:y = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 3 \\ \\ \sf 1 & \sf - 1 \\ \\ \sf 2 & \sf 1 \end{array}} \\ \end{gathered}$

➢ See the attachment, black line represents 2x - y = 3.

Hence,

$\rm :\implies\: \boxed{ \purple{ \bf \: solution \: is \: x \: = 1 \: \: and \: \: y \: = \: - \: 1 }}$