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Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x and y in each system

$x + 2 y = 5$

$2 x - 3 y = - 4$

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X-axis =(5,0), Y-axis=(−2,0)

Y-axis=(−2,0), X-axis =(5,0)

X-axis =(4,0), Y-axis=(−2,0),

Y-axis=(−1,0), X-axis =(4,0),

answer is A.

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Detailed Solution

Given linear equation,

x + 2 y = 5 \dots \dots \dots . . (1)

2 x - 3 y = - 4 \dots \dots \dots \dots . (2)

Put

x = 0

in equation (1),

\Rightarrow 0 + 2 y = 5

\Rightarrow 2 y = 5

\Rightarrow y = \frac{5}{2}

Put

y = 0

in equation (1),

\Rightarrow x + (2 \times 0) = 5

\Rightarrow x + 0 = 5

\Rightarrow x = 5

So, the straight line whose linear equation is

x + 2 y = 5

has two passing points as

A (0, \frac{5}{2})

and

B (5, 0)

.
Substitute

x = 0

in equation

(2)

, we get

\Rightarrow (2 \times 0) - 3 y = - 4

\Rightarrow 0 - 3 y = - 4

\Rightarrow y = \frac{- 4}{- 3}

\Rightarrow y = \frac{4}{3}

Put

y = 0

in equation (2), we get

\Rightarrow 2 x - (3 \times 0) = - 4

\Rightarrow 2 x - 0 = - 4

\Rightarrow x = \frac{- 4}{2}

\Rightarrow x = - 2

So, the straight line whose linear equation is

2 x - 3 y = - 4

have two passing points as

C (0, \frac{4}{3})

and

D (- 2, 0)

.
Equation (1) can be rewritten as

\Rightarrow x = 5 - 2 y \to (3)

Put the value of

x

from equation (3) in equation (2), we have

\Rightarrow 2 (5 - 2 y) - 3 y = - 4

\Rightarrow 10 - 4 y - 3 y = - 4

\Rightarrow 10 - 7 y = - 4

\Rightarrow 7 y = 14

\Rightarrow y = 2

Put

y = 2

in equation (3), we get

\begin{matrix} \Rightarrow x = 5 - (2 \times 2) \\ \Rightarrow x = 5 - 4 \\ \Rightarrow x = 1 \end{matrix}

Now, draw the graph with these points.
Question Image

Here, intersection of lines is

(1, 2)

. So, the solution set is (1,2).
Also, when lines meet the axis then the points are,
X-axis = (5, 0), Y-axis= (−2, 0)
Correct option is 1.

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