Class 9 Mathematics Chapter 13: Two Variables, One Line

It is an equation in two unknowns in which each variable appears to the first power and the variables are not multiplied together, for example $4x + 3y = 24$.

$4x + 3y - 24 = 0$, so $a = 4$, $b = 3$ and $c = -24$.

In $xy = 6$ the two variables are multiplied together, and in $x^2 + y = 5$ the variable $x$ is squared. Linear requires each variable to the first power with no products.

Choose a value and solve for the other each time. With $x = 0$, $y = 11$, giving $(0, 11)$. With $x = 2$, $6 + y = 11$ so $y = 5$, giving $(2, 5)$. With $y = 2$, $3x = 9$ so $x = 3$, giving $(3, 2)$.

For $(3, 4)$ the left side is $4(3) + 3(4) = 12 + 12 = 24$, which matches. For $(4, 3)$ it is $4(4) + 3(3) = 16 + 9 = 25$, which does not. It shows that the order of an ordered pair matters: the first entry is the value of $x$, so swapping the entries gives a different point.

With two variables you may choose any value for one and the other follows, and the choice is unlimited. With one variable there is nothing to choose: $3x = 12$ forces $x = 4$.

The graph is always a straight line, and the line is exactly the set of solutions: every point on it is a solution and every solution lies on it.

Setting $x = 0$ gives $5y = 20$, so $y = 4$ and the point is $(0, 4)$. Setting $y = 0$ gives $2x = 20$, so $x = 10$ and the point is $(10, 0)$. Plot those two and draw the straight line through them, then find a third solution as a check that it lands on the same line.

Here $c = 0$, so setting $x = 0$ gives $y = 0$ and setting $y = 0$ gives $x = 0$ — both attempts return the same point, the origin, and one point cannot fix a line. Choose another value instead: $x = 4$ gives $12 - 4y = 0$, so $y = 3$. The two points $(0, 0)$ and $(4, 3)$ will do.

If both were zero the equation would contain no variable at all, leaving only $c = 0$. That is either a true statement or a false one rather than an equation in two variables.

With two unknowns a single number cannot be an answer, because it would say nothing about the other variable. A solution is therefore a PAIR of values, one for $x$ and one for $y$, which satisfy the equation together, and it is written as an ordered pair $(x, y)$ in which the first entry is always the value of $x$.

To find one, choose any value for either variable and solve the one-variable equation that is left. In $3x + y = 11$, choosing $x = 2$ leaves $6 + y = 11$, so $y = 5$ and the solution is $(2, 5)$. Choosing $x = 0$ leaves $y = 11$, giving $(0, 11)$. You may equally start from $y$: choosing $y = 2$ leaves $3x = 9$, so $x = 3$ and the solution is $(3, 2)$.

Nothing restricted those choices, and nothing ever will. Each new value of $x$ produces its own value of $y$, so the solutions cannot run out: there are infinitely many. This is the real difference from the one-variable equations of earlier grades, where $3x = 12$ pins $x$ to the single value 4.

Plotting the three pairs above shows what the infinity looks like. They do not scatter; they lie in a straight row, and so does every other solution. The graph of the equation is a straight line, and that line is precisely the set of its solutions.

The graph is a straight line, so the task is to find points on it. The easiest are the axis crossings. Setting $x = 0$ gives $5y = 20$, so $y = 4$ and one point is $(0, 4)$. Setting $y = 0$ gives $2x = 20$, so $x = 10$ and the other is $(10, 0)$.

Plot those two points and draw the straight line through them. Two points are enough because a straight line is fixed by any two of its points: there is exactly one line through them, so no further information is needed to know where the line goes.

A third point is still worth finding, not to fix the line but to check the work. Take $x = 5$: then $10 + 5y = 20$, so $y = 2$ and the point is $(5, 2)$. If it lands on the line already drawn, the arithmetic for all three points is consistent. If it misses, one of the three contains a slip, and that is far better discovered now than carried forward.

Finally, every point of the line is a solution, so the drawn line is a complete picture of infinitely many solutions on one page.

As an equation, $4x + 3y = 24$ has infinitely many solutions: choosing any value for $x$ gives a value for $y$. For example $x = 1$ gives $3y = 20$, so $y = \frac{20}{3}$, and $x = 7$ gives $3y = -4$, so $y$ is negative.

As a purchase, most of those are meaningless. You cannot buy a fraction of a notebook, so both values must be whole numbers, and you cannot buy a negative number of pens, so both must be zero or more.

Testing the whole numbers: $x = 0$ gives $y = 8$; $x = 3$ gives $3y = 12$ so $y = 4$; $x = 6$ gives $y = 0$. The other values of $x$ leave $24 - 4x$ not divisible by 3. So the sensible purchases are $(0, 8)$, $(3, 4)$ and $(6, 0)$, or just $(3, 4)$ if at least one of each is wanted.

The lesson is that the graph shows every solution of the equation, while the situation it came from may accept only some of them. The mathematics is not wrong about the others; they simply do not describe a purchase.

When $c = 0$ the line passes through the origin, because substituting $x = 0$ and $y = 0$ satisfies the equation. The practical effect is that the axis-crossing method fails: both crossings are the same point, and one point cannot fix a line. For $3x - 4y = 0$ you choose another value instead, say $x = 4$, which gives $y = 3$ and the second point $(4, 3)$.

When $b = 0$ and $a$ is not, the equation contains no $y$ at all. It fixes $x$ at a single value while $y$ remains free to be anything, so the graph is a line parallel to the $y$-axis — $x = 2$, for instance.

When $a = 0$ and $b$ is not, the mirror happens: $y$ is fixed and $x$ is free, so the graph is a line parallel to the $x$-axis.

If $a$ and $b$ were both zero, no variable would remain. The equation would reduce to $c = 0$, which is simply a true or a false statement about one number and has no graph in the plane at all. That is why the standard form carries the condition that $a$ and $b$ are not both zero.

The slope is the rise divided by the run — the vertical step over the horizontal step between two points. For $(x_1, y_1)$ and $(x_2, y_2)$ it is $m = (y_2 - y_1)/(x_2 - x_1)$.

The rise is $11 - 2 = 9$ and the run is $4 - 1 = 3$, so $m = 9/3 = 3$.

The rise is $-1 - 7 = -8$ and the run is $2 - (-2) = 4$, so $m = -8/4 = -2$. The line falls to the right.

The slope compares the change in $y$ with the change in $x$ for the SAME journey between the two points, so both differences must be taken from the same starting point. If one is reversed, that difference changes sign while the other does not, so the quotient comes out with the opposite sign. A rising line would then appear to fall, which is a wrong description of the line rather than a small slip.

Moving between any two points of a straight line, the rise and the run grow in the same proportion, so their ratio is unchanged. It matters because it makes slope a property of the LINE rather than of the pair of points chosen. That is why a line has exactly one slope and why anyone computing it gets the same answer.

The slope is 3 and the $y$-intercept is $-4$, so the line crosses the $y$-axis at $(0, -4)$. The minus sign belongs to $c$.

Setting $x = 0$ makes the term $mx$ vanish and leaves $y = c$, so the line passes through $(0, c)$, which is exactly where it meets the $y$-axis.

Subtracting $2x$ from both sides gives $y = -2x + 6$. The slope is $-2$, so the line falls two units for every one across, and the intercept is 6, so it crosses the $y$-axis at $(0, 6)$.

$y = 7$ is horizontal through $(0, 7)$. It has no $x$ term, so its slope is 0, and it is already in the form $y = mx + c$ with $m = 0$ and $c = 7$.

$x = -3$ is vertical through $(-3, 0)$. Its run is 0, so its slope is undefined rather than infinite, and it cannot be written as $y = mx + c$ at all, because that form supplies one value of $y$ for each $x$ whereas here $y$ is free and $x$ is fixed.

They are parallel. Their slopes are equal, so they lean at the same rate, but their intercepts differ, so they sit in different places and never meet.

Slope is a number that measures steepness, so that two lines can be compared rather than merely described. Walking along a line from one point to another, the horizontal step is the run and the vertical step is the rise, and the slope is the rise divided by the run.

For two points $(x_1, y_1)$ and $(x_2, y_2)$ this becomes $m = (y_2 - y_1)/(x_2 - x_1)$. Through $P(1, 2)$ and $Q(4, 11)$ the rise is $11 - 2 = 9$ and the run is $4 - 1 = 3$, so $m = 3$: the line climbs three units for every one across. Both subtractions must run in the same order, or the sign flips and a rising line appears to fall.

The sign carries the direction. Through $R(-2, 7)$ and $S(2, -1)$ the rise is $-1 - 7 = -8$ and the run is $2 - (-2) = 4$, so $m = -2$ and the line falls two units for every one across. A positive slope rises to the right and a negative slope falls.

Two values are special. A slope of 0 means no rise over any run, so the line is horizontal. A vertical line has a run of 0, and since division by zero yields no number its slope is undefined — not infinite, because there is no number of any size to name.

Finally, the choice of points does not matter: any two points of the line give the same ratio, which is what makes slope a property of the line itself.

Rearranging a linear equation to make $y$ the subject gives $y = mx + c$, the slope-intercept form. Its value is that both letters are then facts about the line: $m$ is the slope and $c$ is the $y$-intercept.

That $c$ is the intercept follows from the form itself. Setting $x = 0$ makes the term $mx$ vanish and leaves $y = c$, so the line passes through $(0, c)$ — and $(0, c)$ is on the $y$-axis, because every point of that axis has $x = 0$.

To sketch $2x + y = 6$, first rearrange: subtracting $2x$ gives $y = -2x + 6$. So $m = -2$ and $c = 6$.

Now sketch directly. Mark the crossing point $(0, 6)$ on the $y$-axis. The slope is $-2$, meaning the line drops two units for every one unit to the right, so from $(0, 6)$ step one right and two down to reach $(1, 4)$, and rule the line through the two points.

No table of values was needed, because the equation itself gave a point to start from and a direction to travel in. A quick check confirms it: putting $x = 3$ into the original gives $6 + y = 6$, so $y = 0$ and $(3, 0)$ should lie on the drawn line, which it does.

A horizontal line has equation of the form $y = k$. For $y = 7$, there is no $x$ term, so moving along the line produces a run with no rise at all and the slope is 0. It fits the slope-intercept form comfortably, as $y = 0x + 7$, with $m = 0$ and $c = 7$. Every point on it has $y = 7$ while $x$ is free, so it passes through $(0, 7)$, $(5, 7)$ and so on.

A vertical line has equation of the form $x = k$. For $x = -3$, moving along the line produces a rise with a run of 0. The slope would be a division by zero, which yields no number, so the slope is UNDEFINED. It is a common error to call it infinite; there is no number of any size that the ratio approaches.

It also does not fit $y = mx + c$, and the reason is structural rather than a matter of finding the right $m$. That form supplies exactly one value of $y$ for each value of $x$, whereas a vertical line has every value of $y$ at the single value $x = -3$. No choice of $m$ and $c$ can describe that.

So the two cases are not mirror images: one is an easy special case of the general form, and the other falls outside it.

Changing $c$ alone slides the line up or down without altering its lean. The slope is untouched, so the line keeps its direction, but the crossing point moves from $(0, c)$ to the new value. Going from $y = 3x - 4$ to $y = 3x + 2$ lifts the line by six units and leaves it leaning exactly as before.

Changing the sign of $m$ alone keeps the crossing point and reverses the lean. The line still passes through $(0, c)$, but where it rose to the right it now falls at the same rate, as in going from $y = 3x + 2$ to $y = -3x + 2$. The size of $m$ is the steepness and its sign is the direction.

It follows that two lines with equal slopes and different intercepts are parallel. Equal slopes mean they rise at the same rate, so the vertical gap between them never closes; different intercepts mean that gap is not zero to begin with. They therefore never meet, however far they are extended.

And if two equations have the same slope AND the same intercept, they are not two lines at all but one line written twice — a case that matters when a PAIR of equations is solved.

It must satisfy both equations at the same time. Satisfying only one is not enough, because the pair asks for a point lying on both lines.

In the first, $2(3) + 2 = 6 + 2 = 8$. In the second, $3 - 2 = 1$. Both hold, so $(3, 2)$ is a solution of the pair.

The first equation already gives $y$, so substitute into the second: $3x + (2x - 1) = 9$, which is $5x - 1 = 9$, so $5x = 10$ and $x = 2$. Then $y = 2(2) - 1 = 3$. Checking in the second equation: $3(2) + 3 = 9$. The solution is $(2, 3)$.

The $y$ terms are $+y$ and $-y$, so adding the equations removes $y$: $3x = 9$, giving $x = 3$. Substituting into $x - y = 1$ gives $3 - y = 1$, so $y = 2$. Checking in the first: $2(3) + 2 = 8$. The solution is $(3, 2)$.

Each equation is a straight line, and a solution of the pair is a point lying on both, that is a crossing point. Two lines that are not parallel cross at exactly one point.

By scaling: doubling the first equation gives $2x + 4y = 10$. The second demands $2x + 4y = 14$, and the same expression cannot equal both 10 and 14, so no pair can satisfy both.

By ratios: in standard form $a_1/a_2 = 1/2$ and $b_1/b_2 = 2/4 = 1/2$, which agree, while $c_1/c_2 = 5/14$, which differs. Equal coefficient ratios with an unequal constant ratio means the lines are parallel and distinct, so there is no solution.

Doubling the first equation gives exactly $6x + 4y = 24$, which is the second, so the second adds no new condition and both describe the same line. By ratios, $3/6 = 2/4 = 12/24 = 1/2$: all three agree, which is the test for infinitely many solutions. Every point of the single line satisfies both equations.

With both equations as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$: if $a_1/a_2 \neq b_1/b_2$ the lines cross and there is exactly one solution. If $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ they are parallel and distinct, so there is no solution. If all three are equal they are the same line, so there are infinitely many solutions.

That the pair has no solution. The two equations demand something contradictory, so no ordered pair can satisfy both, which means the lines are parallel and distinct.

Let $a$ and $c$ be the prices. Then $2a + 3c = 160$ and $3a + 2c = 165$. Multiplying the first by 3 and the second by 2 gives $6a + 9c = 480$ and $6a + 4c = 330$. Subtracting removes $a$: $5c = 150$, so $c = 30$. Then $2a + 90 = 160$, so $a = 35$. Checking in the second equation: $3(35) + 2(30) = 105 + 60 = 165$.

By substitution. Rearranging the second equation gives $x = y + 1$. Substituting into the first: $2(y + 1) + y = 8$, so $2y + 2 + y = 8$, giving $3y = 6$ and $y = 2$. Then $x = 2 + 1 = 3$.

By elimination. The $y$ terms are already $+y$ and $-y$, so adding the two equations cancels $y$ outright: $(2x + y) + (x - y) = 8 + 1$ gives $3x = 9$ and $x = 3$. Substituting into $x - y = 1$ gives $3 - y = 1$, so $y = 2$.

Both give $(3, 2)$, and the check confirms it in each original equation: $2(3) + 2 = 8$ and $3 - 2 = 1$.

Elimination suited this pair better. The $y$ coefficients were already $+1$ and $-1$, so one addition removed a variable with no rearranging and no fractions at all. Substitution needed an extra rearrangement step first. The general rule the two methods suggest is to use substitution when a variable already stands alone, and elimination when a coefficient already matches or nearly matches.

Each equation is a straight line, and a solution of the pair is a point lying on both lines. So the number of solutions is the number of points the two lines share, and two lines can be arranged in only three ways.

One solution: the lines cross once. $2x + y = 8$ and $x - y = 1$ meet at $(3, 2)$, and two lines that lean differently can meet nowhere else.

No solution: the lines are parallel and distinct, so they share no point. For $x + 2y = 5$ and $2x + 4y = 14$, doubling the first gives $2x + 4y = 10$, and one expression cannot be both 10 and 14.

Infinitely many: the two equations describe the same line, so every one of its points satisfies both. $3x + 2y = 12$ and $6x + 4y = 24$ are such a pair, the second being exactly twice the first and so carrying no new information.

The ratio test tells the cases apart without drawing. With both equations written as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$: $a_1/a_2 \neq b_1/b_2$ gives one solution; $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ gives none; and all three equal gives infinitely many. The first two ratios settle whether the lines lean alike, and the third then settles whether they are one line or two.

(i) Here $a_1/a_2 = 1/2$, $b_1/b_2 = 2/4 = 1/2$ and $c_1/c_2 = 5/14$. The first two agree and the third differs, so the lines are parallel and distinct and the pair has NO SOLUTION.

(ii) Here $a_1/a_2 = 3/6 = 1/2$, $b_1/b_2 = 2/4 = 1/2$ and $c_1/c_2 = 12/24 = 1/2$. All three agree, so the two equations describe the same line and the pair has INFINITELY MANY SOLUTIONS.

(iii) Here $a_1/a_2 = 2/1 = 2$ and $b_1/b_2 = 1/(-1) = -1$. The first two already differ, so the lines lean differently, cross once, and the pair has EXACTLY ONE SOLUTION — and there is no need to look at the third ratio at all once the first two disagree.

Note the order of work: compare the coefficient ratios first, and only when they agree is the constant ratio worth computing.

Let $a$ be the price of an adult ticket and $c$ that of a child ticket, both in rupees. The two purchases give

$2a + 3c = 160$ and $3a + 2c = 165$.

Neither variable stands alone, so elimination is the sensible method. Multiply the first equation by 3 and the second by 2, so that both contain $6a$:

$6a + 9c = 480$ and $6a + 4c = 330$.

Subtracting the second from the first removes $a$ entirely: $5c = 150$, so $c = 30$.

Substituting back into the first original equation: $2a + 3(30) = 160$, so $2a + 90 = 160$, giving $2a = 70$ and $a = 35$.

Check in the OTHER original equation, the one not used for the substitution: $3(35) + 2(30) = 105 + 60 = 165$, which matches. So an adult ticket costs ₹35 and a child ticket ₹30.

It is an equation in two unknowns in which each variable appears to the first power and the variables are not multiplied together, for example $4x + 3y = 24$.

$4x + 3y - 24 = 0$, so $a = 4$, $b = 3$ and $c = -24$.

In $xy = 6$ the two variables are multiplied together, and in $x^2 + y = 5$ the variable $x$ is squared. Linear requires each variable to the first power with no products.

The slope is the rise divided by the run — the vertical step over the horizontal step between two points. For $(x_1, y_1)$ and $(x_2, y_2)$ it is $m = (y_2 - y_1)/(x_2 - x_1)$.

The rise is $11 - 2 = 9$ and the run is $4 - 1 = 3$, so $m = 9/3 = 3$.

The rise is $-1 - 7 = -8$ and the run is $2 - (-2) = 4$, so $m = -8/4 = -2$. The line falls to the right.

It must satisfy both equations at the same time. Satisfying only one is not enough, because the pair asks for a point lying on both lines.

In the first, $2(3) + 2 = 6 + 2 = 8$. In the second, $3 - 2 = 1$. Both hold, so $(3, 2)$ is a solution of the pair.

The first equation already gives $y$, so substitute into the second: $3x + (2x - 1) = 9$, which is $5x - 1 = 9$, so $5x = 10$ and $x = 2$. Then $y = 2(2) - 1 = 3$. Checking in the second equation: $3(2) + 3 = 9$. The solution is $(2, 3)$.