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Ml aggarwal solutions

. Solve graphically the following equations: x + 2y = 4, 3x – 2y = 4

Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the x-axis.

x+2y = 4 …(i)

2y = 4-x

y = (4-x)/2

When x = 0, y = (4-0)/2 = 4/2 = 2

When x = 2, y = (4-2)/2 = 2/2 = 1

When x = 4, y = (4-4)/2 = 0/2 = 0

3x-2y = 4 ..(ii)

2y = 3x-4

y = (3x-4)/2

When x = 0, y = (3×0-4)/2 = (0-4)/2 = -4/2 = -2

When x = 2, y = (3×2-4)/2 = (6-4)/2 = 2/2 = 1

When x = 4, y = (3×4-4)/2 = (12-4)/2 = 8/2 = 4

It is clear from the graph that the two lines intersect at (2,1).

So the solution of the given equations is x = 2 and y = 1.

The area of the triangle formed by these lines and X-axis = ½ ×base×height

= ½ ×2.7×1

= 1.35 sq. units

Hence the area of the triangle is 1.35 sq. units

. Solve graphically the following equations: x + 2y = 4, 3x – 2y = 4

Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the x-axis.

x+2y = 4 …(i)

2y = 4-x

y = (4-x)/2

When x = 0, y = (4-0)/2 = 4/2 = 2

When x = 2, y = (4-2)/2 = 2/2 = 1

When x = 4, y = (4-4)/2 = 0/2 = 0

3x-2y = 4 ..(ii)

2y = 3x-4

y = (3x-4)/2

When x = 0, y = (3×0-4)/2 = (0-4)/2 = -4/2 = -2

When x = 2, y = (3×2-4)/2 = (6-4)/2 = 2/2 = 1

When x = 4, y = (3×4-4)/2 = (12-4)/2 = 8/2 = 4

It is clear from the graph that the two lines intersect at (2,1).

So the solution of the given equations is x = 2 and y = 1.

The area of the triangle formed by these lines and X-axis = ½ ×base×height

= ½ ×2.7×1

= 1.35 sq. units

Hence the area of the triangle is 1.35 sq. units

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