Question

solve the following system of equations graphically. x + 3y = 6 2x-3y=12​

Answer 1

Given Equation

x + 3y = 6 (i)

2x - 3y = 12 (ii)

Now Take (i) eq

x + 3y = 6

when x = 0

0 + 3y = 6

3y = 6

y = 6/3

y = 2

We get point (0,2)

when y = 0

x + 3(0) = 6

x = 6

we get point (6,0)

we get point (6,0)Now take (ii) eq

2x - 3y = 12

when x = 0

2(0) - 3y = 12

-3y = 12

y = -12/3

y = -4

we get point (0,-4)

we get point (0,-4)when y =0

2x - 3(0) = 12

2x = 12

x = 12/2

x = 6

we get point (6, 0)

Answer 2

DIAGRAM:-

$\setlength{\unitlength}{.5in}\begin{picture}(0,0)\thinlines\multiput(-2,-2)(0.5,0){30}{\line(0,1){20}}\multiput(-2,-2)(0,0.5){20}{\line(1,0){21}}\thicklines\put (0,-1){\vector (0,1){8}}\put (0,-1){\vector (0,-1){0.1}}\put (-1,0){\vector(1,0){13}}\put (-1,0){\vector (-1,0){0.1}}\put (-1.2,-0.4){\sf X'}\put (12.25,-0.4){\sf X}\put (-0.2,-1.4){\sf Y'}\put (-0.2,9){\sf Y}\put(6,0){\circle*{0.3}}\put(9,2){\circle*{0.3}}\put(3,-2){\circle*{0.3}}\put(3,1){\circle*{0.3}}\put(0,2){\circle*{0.3}}\put(6,0){\vector(-3,1){8}}\put(6,0){\vector(3,-1){4}}\put(6,0){\vector(3,2){5}}\put(6,0){\vector(-3,-2){5}}\end{picture}$

$\sf x+3y=6\dots(1)$

$\sf 2x-3y=12\dots(2)$

From eq(1)

$\\ \sf{:}\longrightarrow x+3y=6$

$\\ \sf{:}\longrightarrow x=6-3y$

Table:-

$\boxed{\begin{array}{c|c|c|c}\bf x &\rm 6 &\rm 3 &\rm 0 \\ \bf y &\rm 0 &\rm 1 &\rm 2 \end{array}}$

we got points as

• (6,0)
• (3,1)
• (0,2)

From eq(2)

$\\ \sf{:}\longrightarrow 2x-3y=12$

$\\ \sf{:}\longrightarrow 2x=12+3y$

$\\ \sf{:}\longrightarrow x=\dfrac{12+3y}{2}$

Table:-

$\boxed{\begin{array}{c|c|c|c}\bf x &\rm 6 &\rm 9 &\rm 3 \\\bf y & \rm 0 &\rm 2 &\rm -2 \end{array}}$

we got points as

• (6,0)
• (9,2)
• (3,-2)

Hence the solution is (6,0)

Note:-

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