Linear Equations in Two Variables — NCERT Solutions, AI Tutor & Practice

Start with the given equation:

$3x + 7 = 5y$

Bring all terms to the left-hand side:

$3x - 5y + 7 = 0$

Comparing with the standard form $ax + by + c = 0$:

• $a = 3$
• $b = -5$
• $c = 7$

Given:

$y = -4x$

Move all terms to the left-hand side:

$4x + y = 0$

Writing the constant term explicitly:

$4x + y + 0 = 0$

Comparing with $ax + by + c = 0$:

• $a = 4$
• $b = 1$
• $c = 0$

Let the cost of a notebook be $\rupee\, x$ and the cost of a pen be $\rupee\, y$.

The statement says the notebook costs twice the pen:

$x = 2y$

Writing in standard form:

$x - 2y = 0$

This is the required linear equation in two variables, with $a = 1$, $b = -2$, $c = 0$.

For $2x + y = 7$, choose values of $x$ and compute $y = 7 - 2x$.

• $x = 0 \Rightarrow y = 7 - 0 = 7 \Rightarrow (0, 7)$
• $x = 1 \Rightarrow y = 7 - 2 = 5 \Rightarrow (1, 5)$
• $x = 2 \Rightarrow y = 7 - 4 = 3 \Rightarrow (2, 3)$
• $x = 3 \Rightarrow y = 7 - 6 = 1 \Rightarrow (3, 1)$

Four solutions are: $(0,7),\ (1,5),\ (2,3),\ (3,1)$.

Since infinitely many values of $x$ are possible, the equation has infinitely many solutions.

Substitute each point into $4x + 3y$ and compare with $11$.

For $(2, 1)$:

$4(2) + 3(1) = 8 + 3 = 11$

This equals the RHS, so $(2, 1)$ is a solution.

For $(-1, 5)$:

$4(-1) + 3(5) = -4 + 15 = 11$

This also equals $11$, so $(-1, 5)$ is a solution.

Both ordered pairs are solutions of $4x + 3y = 11$.

Since $(3, k)$ is a solution, substitute $x = 3$ and $y = k$:

$2(3) - 5k = 1$

$6 - 5k = 1$

$-5k = 1 - 6 = -5$

$k = \frac{-5}{-5} = 1$

Therefore $k = 1$, and the solution point is $(3, 1)$.

Find at least two solutions of $x + y = 5$ (using $y = 5 - x$):

| $x$ | $y = 5 - x$ | Point | |-----|------------|-------| | $0$ | $5$ | $(0, 5)$ | | $5$ | $0$ | $(5, 0)$ | | $2$ | $3$ | $(2, 3)$ |

Plotting: Mark the points $(0,5)$, $(5,0)$ and $(2,3)$ on the coordinate plane.

Line: Since all three points lie in a straight line, join them and extend both ways. This straight line is the graph of $x + y = 5$.

The line meets the $y$-axis at $(0,5)$ and the $x$-axis at $(5,0)$.

Rewrite as $y = 2x - 4$ and prepare a table:

| $x$ | $y = 2x - 4$ | Point | |-----|-------------|-------| | $0$ | $-4$ | $(0, -4)$ | | $2$ | $0$ | $(2, 0)$ | | $3$ | $2$ | $(3, 2)$ |

Plot the points $(0,-4)$, $(2,0)$, $(3,2)$ and join them with a straight line.

Intercepts:

• Cuts the $x$-axis where $y = 0$: at $(2, 0)$.
• Cuts the $y$-axis where $x = 0$: at $(0, -4)$.

Prepare a table of solutions for $y = 3x$:

| $x$ | $y = 3x$ | Point | |-----|----------|-------| | $0$ | $0$ | $(0, 0)$ | | $1$ | $3$ | $(1, 3)$ | | $-1$ | $-3$ | $(-1, -3)$ |

Plot the points and join them to get a straight line.

Verification: When $x = 0$, $y = 3(0) = 0$, giving the point $(0,0)$, which is the origin. Since the origin satisfies the equation, the graph passes through the origin.

Every equation of the form $y = mx$ (with $c = 0$) passes through the origin.

A line parallel to the $x$-axis has every point with the same $y$-coordinate. Its equation is of the form $y = k$.

Since the line passes through $(2, -3)$, the constant $y$-value is $-3$:

$y = -3$

Graph: Draw a horizontal straight line crossing the $y$-axis at $(0, -3)$ and running parallel to the $x$-axis. It passes through points such as $(0,-3)$, $(2,-3)$ and $(-4,-3)$.

The equation $x = 4$ can be written in two variables as $x + 0\cdot y = 4$. Here $x$ is always $4$, while $y$ can be any real number.

Some solutions:

| $x$ | $y$ | Point | |-----|-----|-------| | $4$ | $0$ | $(4, 0)$ | | $4$ | $2$ | $(4, 2)$ | | $4$ | $-1$ | $(4, -1)$ |

Plot these points and join them — they form a vertical straight line parallel to the $y$-axis, cutting the $x$-axis at $(4, 0)$.

Such a line ($x = k$) is always parallel to the $y$-axis.

Solve for $y$:

$2y + 5 = 0 \Rightarrow 2y = -5 \Rightarrow y = -\tfrac{5}{2}$

(i) In one variable: On the number line, $y = -\tfrac{5}{2}$ is a single point located at $-2.5$.

(ii) In two variables: Written as $0\cdot x + 2y + 5 = 0$, the value $y = -\tfrac{5}{2}$ holds for every value of $x$. This is a horizontal straight line parallel to the $x$-axis, passing through $(0, -\tfrac{5}{2})$.

So the same equation represents a point on a line, but a whole line in the plane.