Question

Q.7: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?
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Answer 1

Given:

3x+4y = 6

To find:

At what points, the graph cuts X-axis and Y-axis

Solution:

The equation of the x-axis is y=0 so if you want to get the intersection point of the line on the x-axis, then put the value of y as 0

• 3x+4y = 6
• 3x + 0 = 6
• x = 2

So the point at which the intersect x-axis is (2,0)

The equation of the y-axis is x=0 so if you want to get the intersection point of the line on the y-axis, then put the value of x as 0

• 3x+4y = 6
• 0 + 4y = 6
• y = 3/2

So the point at which the intersect y-axis is (0, 3/2)

(2,0) and (0, 3/2) are the points at which the graph cuts X-axis and Y-axis

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

• To Draw the graph of the linear equation 3x + 4y = 6
• The points where the graph cuts X and Y-axis

CALCULATION

The equation of the given linear equation is

$\sf{3x + 4y = 6}$

Which can be rewritten as below

$\displaystyle \sf{ \frac{3x}{6} + \frac{4y}{6} = 1 \: }$

$\implies \: \displaystyle \sf{ \frac{x}{2} + \frac{y}{ \frac{3}{2} } = 1 \: } \: \: ....(1)$

Which is of the intercept form

$\sf{Thus \: the \: equation \: cuts \: x \: axis \: at \: \: ( 2, 0) \: \: and}$

$\displaystyle \sf{ Y \: axis \: at \: \: ( \: 0 \: , \: \frac{3}{2} \: ) \: }$

GRAPH : The graph of the linear equation is referred to the attachment

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