Linear Equations in Two Variables Test - 16

2x – 5y = 7 is a linear equation in two variables. A linear equation in two variables has infinitely many solutions.

In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation.

2 (2) + 3 (0) = k

4 + 0 = k

k = 4

Hence, the value of k is 4.

The given linear equation is

2x + 0y + 9 = 0

2x + 9 = 0

2x = -9

x = \(-{9\over2}\) and y can be any real number.

Hence, (\(-{9\over2}\) , m) is the required form of solution of the given linear equation.

Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis.

So, we put x = 0 in the given equation 2x+ 3y = 6, we get

2 x 0 + 3y = 6

3y = 6

y = 2.

Hence, at the point (0, 2), the given linear equation cuts the Y-axis.

Here, the’Coefficient of y in the given equation x =7 is 0. So, the equation can be written as

\(1 \times x + 0 \times y = 7\)

Hence, the required equation is 1.x + 0. y = 7.

Every point on the X-axis has its y-coordinate equal to zero. i.e., y=0

Hence, the general form of every point on X-axis is (x, 0).

Any point on the line y = x will have x and y coordinate same.

So, any point on the line y = x is of the form (a, a).

y = 0 is the equation of x-axis.

The given equation y = 6 does not contain x. Its graph is a line parallel to x-axis. So, the graph of y = 6 is a line parallel to x-axis at a distance 6 units from the origin.

x = 5, y = 2 is a solution of the linear equation x + y = 7, as 5 + 2 = 7.

The points (-2, 2) and (2, -2) have x and y coordinates of opposite signs.

Also, any point on the graph of x + y = 0

i.e., y = -x will have x and y coordinate of opposite signs. The Point (0, 0) also satisfies

x + y = 0.

We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

Since, the graph of linear equation 2x + 3y = 6 meets the X-axis.

So, we put y = 0 in 2x + 3y = 6

2x + 3(0) = 6

2x + 0 = 6

x = \(6\over2\) 

x = 3

Hence, the coordinate on X-axis is (3, 0).

The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same.

There are infinitely many linear equations which are satisfied by x = 1 and y = 2.

For example, a linear equation x + y = 3 is satisfied by x = 1 and y =2.

Others are y = 2x, y – x = 1, 2y – x = 3 etc.

Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)]

Hence, the point (a, – a) always lies on the line x + y = 0.

Here, you can see, the distance is 4 units. The distance can be calculated by subtracting y coordinates. 3-(-1) =4.