Pair of Linear Equations in Two Variables

Akhila goes to a village fair. She rides the Giant Wheel some number of times and plays Hoopla — tossing rings onto prizes — some other number of times. She knows two things: the number of times she played Hoopla is half the number of Giant Wheel rides, and with rides at ₹3 each and Hoopla at ₹4 a turn, she spent exactly ₹20 in total. Two facts, two unknowns, and no way to pin either one down from just one of the facts alone. This is the puzzle that opens NCERT's Class 10 chapter on the pair of linear equations in two variables, and it is really the puzzle behind an enormous range of everyday problems — two ages, two digits of a number, two speeds, two prices — that only become solvable once you write down two equations connecting two unknowns together. You already know, from Class 9, that a single linear equation like $3x + 4y = 20$ has infinitely many solutions, one for every point on the straight line it draws. This chapter asks: what happens when two such equations must hold simultaneously? Geometrically, each equation is a line, and asking for a common solution is asking where the two lines meet. Three things can happen — the lines can cross once, run parallel forever, or lie exactly on top of each other — and this chapter builds the complete toolkit for telling which case you are in and for finding the exact answer. You will meet the graphical method, which turns solving into plotting; a purely algebraic ratio test that predicts the outcome from the coefficients alone, without drawing anything; and two efficient algebraic techniques, the substitution method and the elimination method, that solve the pair exactly, including for answers that are not neat whole numbers a graph could never show precisely. By the end, Akhila's fair — and dozens of problems just like it — become two lines on a page and a handful of clean algebraic steps.

Key takeaways

A linear equation in two variables has the general form $ax + by + c = 0$ , with $a$ and $b$ not both zero; on its own it has infinitely many solutions, each one a point on a straight line.
A pair of linear equations in two variables is two such equations, $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ , considered together; a common solution $(x,y)$ must satisfy both at once.
Graphically, the two lines can intersect at one point (unique solution, a consistent pair), be parallel (no solution, an inconsistent pair), or coincide (infinitely many solutions, a dependent pair — always consistent).
Comparing the ratios $\frac{a_1}{a_2}$ , $\frac{b_1}{b_2}$ , $\frac{c_1}{c_2}$ predicts which of the three cases holds without drawing a single graph.
The graphical method is intuitive but breaks down when the true solution has non-integral or irrational coordinates that cannot be read accurately off a graph.
The substitution method isolates one variable from one equation and substitutes that expression into the other equation, collapsing the pair into a single equation in one variable.
The elimination method scales the two equations so that one variable's coefficients match, then adds or subtracts the equations to remove that variable entirely.
If elimination or substitution collapses the pair into a false statement with no variable left (like $0 = 9$ ), the pair has no solution; if it collapses into a true statement (like $18 = 18$ ), the pair has infinitely many solutions.
Many word problems — ages, two-digit numbers, ticket prices, taxi fares, speeds — translate directly into a pair of linear equations, and the answer should always be checked against the original conditions of the problem.
Key idea: a pair of linear equations a1x+b1y+c1=0 and a2x+b2y+c2=0 represents two lines whose relative position (intersecting, parallel or coincident) is fixed entirely by comparing a1/a2, b1/b2 and c1/c2, and the exact solution, when it exists and is unique, is always recoverable by substitution or elimination.

The core insight

A pair of linear equations in two variables describes two straight lines, and comparing the ratios of their coefficients ( $a_1/a_2$ against $b_1/b_2$ against $c_1/c_2$ ) tells you instantly whether they intersect once, never meet, or coincide entirely — while substitution and elimination give the exact point of intersection whenever one exists.

What this chapter covers

The chapter has two large halves. The first half is graphical: it defines what a pair of linear equations in two variables is, shows that solving the pair means finding where two lines meet, and sorts every possible pair into exactly three cases — intersecting, parallel, or coincident lines — with the matching language of consistent, inconsistent and dependent pairs. It then shows that these three cases can be read off directly from the coefficients, by comparing the ratios $a_1/a_2$ , $b_1/b_2$ and $c_1/c_2$ , without plotting anything. The second half is algebraic, and answers the natural question: what do you do when a graph cannot give you an accurate answer? It develops two systematic methods that always work whenever a unique solution exists — the substitution method, which reduces the pair to a single equation by replacing one variable with an expression in the other, and the elimination method, which scales and combines the two equations to cancel one variable outright. Throughout, the chapter roots every technique in word problems — fair rides, incomes and expenditures, ages, two-digit numbers — so that forming the pair of equations from a real situation is as much a skill as solving it.

The core ideas

Linear equations in two variables, and a pair of them

A single equation of the form $ax + by + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a$ , $b$ are not both zero, is a linear equation in two variables. Its graph is always a straight line, and every point on that line is one of its infinitely many solutions. This chapter studies what happens when two such equations are imposed together on the same pair of unknowns $x$ and $y$ :

a_1x + b_1y + c_1 = 0, \qquad a_2x + b_2y + c_2 = 0. \qquad (1)

where: $a_1, b_1, c_1$ are the coefficients and constant of the first equation, $a_2, b_2, c_2$ those of the second, and a solution of the pair is any $(x, y)$ that satisfies both equations at once — geometrically, a point that lies on both of the two lines simultaneously.

The three graphical possibilities

Because each equation of the pair draws a straight line, finding a common solution is exactly the problem of finding where two lines meet. Two distinct straight lines in a plane can be positioned relative to each other in only three ways, and each way has its own name and its own count of solutions. If the lines intersect at a single point, that point is the unique solution, and the pair is called consistent. If the lines are parallel and never meet, the pair has no solution at all, and is called inconsistent. If the lines coincide — they are really the same line drawn twice — every point on that line is a shared solution, giving infinitely many solutions; such a pair is called dependent, and a dependent pair is always consistent, since it certainly does have solutions, just not a unique one.

Figure

Three panels showing a pair of lines that intersect at one point a pair of parallel lines that never meet and a pair of coincident lines lying on top of each other

Solving graphically: NCERT's own example

To see the graphical method at work, take the pair $x + 3y = 6$ and $2x - 3y = 12$ . Two points fix each line: for the first, $(0, 2)$ and $(6, 0)$ ; for the second, $(0, -4)$ and $(3, -2)$ . Plotting and joining these gives two lines, $AB$ and $PQ$ , and they turn out to cross at the point $B(6, 0)$ — so $x = 6$ , $y = 0$ is the unique solution, and the pair is consistent. This is exactly the working of Example 1 of the chapter, and it is worth checking directly by substitution: $6 + 3(0) = 6$ and $2(6) - 3(0) = 12$ , both true.

Figure

A coordinate grid showing the lines x plus three y equals six and two x minus three y equals twelve crossing at the point six comma zero

The graphical method is intuitive, but it has a real limitation: it is not convenient when the true solution has non-integral coordinates like $(\sqrt{3}, 2\sqrt{7})$ or $\left(-1.75, 3\tfrac{3}{10}\right)$ , since reading such coordinates accurately off a hand-drawn graph invites mistakes. This is the motivation for the algebraic methods that follow.

The algebraic ratio test

Rather than plotting, the three cases above can be read straight off the six coefficients of the pair $a_1x+b_1y+c_1=0$ , $a_2x+b_2y+c_2=0$ , by comparing three ratios. Each of the following statements matches one of the graphical outcomes exactly, and each is proved from first principles in the first derivation below.

\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \quad\Longrightarrow\quad \text{intersecting lines: a unique solution (consistent pair).} \qquad (2)

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \quad\Longrightarrow\quad \text{coincident lines: infinitely many solutions (dependent, consistent pair).} \qquad (3)

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \quad\Longrightarrow\quad \text{parallel lines: no solution (inconsistent pair).} \qquad (4)

where: the ratios compare corresponding coefficients of $x$ , of $y$ , and the constant terms across the two equations. Notice the three cases are exhaustive and mutually exclusive — for any pair of linear equations, exactly one of (2), (3), (4) must hold, so this single comparison always settles the question. For instance, $2x+3y-9=0$ and $4x+6y-18=0$ give $\frac{2}{4}=\frac{3}{6}=\frac{-9}{-18}=\frac12$ , matching (3): the lines are coincident, exactly as the chapter's Table 3.1 records.

Substitution method

The substitution method turns the pair into a single equation in one variable in three moves. First, pick either equation and write one variable in terms of the other — say, solve for $x$ in terms of $y$ from the second equation. Second, substitute that expression for $x$ into the first equation wherever $x$ appears; because the first equation now contains only $y$ , it can be solved directly, and this is the step where a true statement with no variable signals infinitely many solutions, and a false one signals no solution. Third, having found $y$ (or $x$ ), substitute it back into the expression from the first step to recover the other variable. A quick illustration: for $x + 2y = 3$ , solving the second equation for $x$ gives $x = 3 - 2y$ ; substituting this into $7x - 15y = 2$ gives a single equation purely in $y$ , which is exactly Example 4's setup.

Elimination method

The elimination method removes one variable by combining scaled copies of the two equations, and is often faster when neither equation is naturally solved for a single variable. First, multiply the two equations by suitable non-zero constants so that the coefficients of one chosen variable — say $y$ — become numerically equal in both. Second, add or subtract the two scaled equations so that variable cancels, leaving a single equation in the other variable, exactly as in the false-statement and true-statement cases noted above. Third, solve that equation. Fourth, substitute the value found back into either of the original equations to get the remaining variable. The chapter's own incomes-and-expenditures problem (Example 8) is the model case, worked in full in the worked examples below.

Derivation AWhy comparing the coefficient ratios predicts intersecting, parallel or coincident lines

The goal of this derivation is to prove, rather than merely state, the three ratio conditions given above — to show exactly why $a_1/a_2 \neq b_1/b_2$ forces the lines to cross, why equal ratios of $a$ and $b$ but unequal ratios of $c$ forces them to run parallel forever, and why all three ratios equal forces them onto the very same line. The proof uses nothing beyond rewriting each equation in slope-intercept form and comparing slopes and intercepts directly, and it is exactly what justifies Table 3.1 of the chapter for every possible pair of coefficients, not just the three sample pairs the chapter lists.

Step 1. Rearrange each equation into slope-intercept form, assuming for this step that $b_1 \neq 0$ and $b_2 \neq 0$ (so neither line is vertical). From $a_1x+b_1y+c_1=0$ , isolating $y$ gives $y = -\dfrac{a_1}{b_1}x - \dfrac{c_1}{b_1}$ , a line with slope $m_1 = -\dfrac{a_1}{b_1}$ and y-intercept $k_1=-\dfrac{c_1}{b_1}$ . Similarly the second equation gives slope $m_2=-\dfrac{a_2}{b_2}$ and intercept $k_2=-\dfrac{c_2}{b_2}$ .

Step 2. Two distinct straight lines intersect at exactly one point precisely when their slopes differ, since equal slopes force them to run in the same direction forever without converging. So the lines intersect exactly when $m_1 \neq m_2$ , that is $-\dfrac{a_1}{b_1} \neq -\dfrac{a_2}{b_2}$ . Cross-multiplying (valid since $b_1,b_2\neq0$ ) turns this into $a_1b_2 \neq a_2b_1$ , which is the same statement as $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$ — exactly the intersecting-lines condition.

Step 3. If instead the slopes are equal, $m_1=m_2$ , so $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ , the two lines are either parallel and distinct or the same line, and the intercepts $k_1$ , $k_2$ decide which. If the intercepts differ ( $k_1 \neq k_2$ ), the lines are parallel but never meet, giving no solution; writing this out, $-\dfrac{c_1}{b_1} \neq -\dfrac{c_2}{b_2}$ rearranges to $\dfrac{c_1}{c_2} \neq \dfrac{b_1}{b_2}$ using the same cross-multiplication idea, which combined with Step 2's equality gives exactly $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$ , the parallel-lines condition.

Step 4. If both the slopes and the intercepts agree ( $m_1=m_2$ and $k_1=k_2$ ), the two equations describe the identical line, so every one of its infinitely many points is a shared solution; this is precisely $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ , the coincident-lines condition. The three cases in Steps 2 to 4 are exhaustive — for any two lines, exactly one of "slopes differ," "slopes equal but intercepts differ," or "both equal" must be true — so this single three-way comparison always classifies the pair correctly, matching every row of Table 3.1, including the coincident pair $2x+3y-9=0$ , $4x+6y-18=0$ checked earlier.

Go deeper · Derivation B · 1 minThe general elimination formula, and why it always solves an intersecting pair

Derivation BThe general elimination formula, and why it always solves an intersecting pair

The goal here is to run the elimination method once in full generality, on the pair $a_1x+b_1y+c_1=0$ , $a_2x+b_2y+c_2=0$ with no numbers substituted yet, so as to obtain explicit formulas for $x$ and $y$ in terms of the six coefficients alone, valid exactly when $a_1b_2-a_2b_1\neq0$ — which, by the previous derivation, is exactly the condition $a_1/a_2\neq b_1/b_2$ for a unique solution. This shows the elimination method is not a trick that happens to work on textbook examples, but a procedure guaranteed to succeed on every consistent pair, and the result is checked numerically against Example 8 in the last step.

Step 1. Write the equations as $a_1x+b_1y=-c_1$ and $a_2x+b_2y=-c_2$ , moving the constants to the right so the elimination arithmetic is easier to track, and assume $a_1b_2-a_2b_1\neq0$ throughout.

Step 2. Eliminate $y$ first. Multiply the first equation by $b_2$ and the second by $b_1$ , giving $a_1b_2x+b_1b_2y=-c_1b_2$ and $a_2b_1x+b_1b_2y=-c_2b_1$ . Subtracting the second from the first cancels the $y$ -terms exactly, leaving $(a_1b_2-a_2b_1)x = b_1c_2-b_2c_1$ , so $x=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$ .

Step 3. Eliminate $x$ next, by the mirror procedure. Multiply the first equation by $a_2$ and the second by $a_1$ , giving $a_1a_2x+a_2b_1y=-a_2c_1$ and $a_1a_2x+a_1b_2y=-a_1c_2$ . Subtracting the first from the second cancels the $x$ -terms, leaving $(a_1b_2-a_2b_1)y = a_2c_1-a_1c_2$ , so $y=\dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}$ .

Step 4. Verify against Example 8, $9x-4y-2000=0$ and $7x-3y-2000=0$ , so $a_1=9,b_1=-4,c_1=-2000$ and $a_2=7,b_2=-3,c_2=-2000$ . The denominator is $a_1b_2-a_2b_1=9(-3)-7(-4)=-27+28=1$ . Then $x=\dfrac{b_1c_2-b_2c_1}{1}=(-4)(-2000)-(-3)(-2000)=8000-6000=2000$ , and $y=\dfrac{a_2c_1-a_1c_2}{1}=7(-2000)-9(-2000)=-14000-(-18000)=4000$ . Both match the elimination-by-scaling answer found step by step in the worked examples below, confirming the general formula reproduces the textbook's own numbers exactly.

Worked examples

Board level - Akhila's fair: forming and solving the pair by substitution

At a village fair, Akhila played Hoopla half as many times as she rode the Giant Wheel. Each ride costs ₹3 and each game of Hoopla costs ₹4, and she spent ₹20 in total. Find the number of rides she had and the number of times she played Hoopla.

Step 1. Name the unknowns and translate both facts into equations. Let $x$ be the number of Giant Wheel rides and $y$ the number of times she played Hoopla. "Hoopla is half the rides" gives $y = \dfrac{1}{2}x$ , and "total spend is ₹20" gives $3x + 4y = 20$ .

Step 2. Substitute the first equation into the second, replacing $y$ with $\tfrac{1}{2}x$ : $3x + 4\left(\dfrac{1}{2}x\right) = 20$ , which simplifies to $3x + 2x = 20$ .

Step 3. Solve the single-variable equation: $5x = 20$ , so $x = 4$ .

Step 4. Substitute $x = 4$ back into $y = \tfrac12 x$ to get $y = 2$ . Check: $3(4) + 4(2) = 12 + 8 = 20$ , correct. Answer: Akhila had 4 rides on the Giant Wheel and played Hoopla 2 times.

Board level - elimination method for two incomes (Example 8 pattern)

The ratio of the monthly incomes of two persons is $9:7$ , and the ratio of their monthly expenditures is $4:3$ . If each of them saves ₹2000 per month, find their monthly incomes.

Step 1. Let the incomes be ₹ $9x$ and ₹ $7x$ , and the expenditures be ₹ $4y$ and ₹ $3y$ . Since savings equal income minus expenditure and both save ₹2000, this gives the pair $9x - 4y = 2000$ and $7x - 3y = 2000$ .

Step 2. To eliminate $y$ , make its coefficients numerically equal: multiply the first equation by $3$ and the second by $4$ , giving $27x - 12y = 6000$ and $28x - 12y = 8000$ .

Step 3. Subtract the first new equation from the second to eliminate $y$ : $(28x - 27x) + (-12y + 12y) = 8000 - 6000$ , so $x = 2000$ .

Step 4. Substitute $x = 2000$ into $9x - 4y = 2000$ : $18000 - 4y = 2000$ , so $4y = 16000$ and $y = 4000$ . The incomes are $9x = ₹18{,}000$ and $7x = ₹14{,}000$ . Check: $18000:14000 = 9:7$ , and the expenditures $16000:12000 = 4:3$ , both correct. Answer: the monthly incomes are ₹18,000 and ₹14,000.

Board HOTS / foundation JEE-NEET - the value of k for which a pair has no solution

Find the value of $k$ for which the pair of linear equations $kx + 3y = k - 3$ and $12x + ky = k$ has no solution.

Step 1. Write both equations in the standard form $ax+by+c=0$ : $kx + 3y - (k-3) = 0$ gives $a_1=k$ , $b_1=3$ , $c_1 = 3-k$ ; and $12x + ky - k = 0$ gives $a_2=12$ , $b_2=k$ , $c_2=-k$ .

Step 2. A pair has no solution exactly when $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ . Start with the first equality: $\dfrac{k}{12} = \dfrac{3}{k}$ , which cross-multiplies to $k^2 = 36$ , so $k = 6$ or $k = -6$ .

Step 3. Both candidate values satisfy $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ by construction, but only one can also satisfy the strict inequality against $\dfrac{c_1}{c_2}$ — so each must be tested separately. For $k=6$ : $\dfrac{a_1}{a_2}=\dfrac{6}{12}=\dfrac12$ , $\dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac12$ , and $\dfrac{c_1}{c_2}=\dfrac{3-6}{-6}=\dfrac{-3}{-6}=\dfrac12$ — all three ratios are equal, so $k=6$ makes the lines coincident (infinitely many solutions), not the case wanted.

Step 4. For $k=-6$ : $\dfrac{a_1}{a_2}=\dfrac{-6}{12}=-\dfrac12$ , $\dfrac{b_1}{b_2}=\dfrac{3}{-6}=-\dfrac12$ , equal as required, but $\dfrac{c_1}{c_2}=\dfrac{3-(-6)}{6}=\dfrac{9}{6}=\dfrac32$ , which is not equal to $-\tfrac12$ . This satisfies $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$ exactly. Answer: k = -6. Timing note: budget three to four minutes. The trap is stopping as soon as $k^2=36$ gives two candidates — the condition for no solution is a strict inequality on the third ratio, so every candidate value from the first equality must be individually tested against $c_1/c_2$ , and roots that make all three ratios equal (like $k=6$ here) must be rejected. This same three-ratio check, applied with the opposite inequality, is exactly how JEE-style problems ask for the value of $k$ that makes a pair dependent or consistent instead.

Common traps

The most damaging confusion in this chapter is treating "no solution" and "infinitely many solutions" as the same kind of failure — an inconsistent pair (parallel lines) genuinely has no answer, while a dependent pair (coincident lines) has every point on the line as an answer; a dependent pair is always consistent, an inconsistent pair never is. A second common slip is a sign error while isolating a variable for substitution, especially when the coefficient being divided out is negative. A third is scaling only one side of an equation during elimination — remember that the constant term must be multiplied by the same factor as the coefficients, since it is still part of the same equation. A fourth is misreading the ratio test by comparing $a_1/b_1$ to $a_2/b_2$ instead of $a_1/a_2$ to $b_1/b_2$ ; get the pairing backwards and the conclusion flips. A fifth, specific to word problems, is forgetting to check the answer against every condition stated in the problem — Example 10's two-digit-number puzzle, for instance, genuinely has two different valid answers because the condition "digits differ by 2" splits into two cases, and missing one loses marks. Finally, students often assume every pair of linear equations must be solvable; verifying with the ratio test before grinding through algebra saves time and catches inconsistent or dependent pairs immediately.

Common trapInconsistent (no solution) and dependent (infinitely many solutions) are opposite outcomes, not variations on the same theme

It is tempting to lump together every pair of equations that fails to give one neat answer, but "no solution" and "infinitely many solutions" are opposite ends of the spectrum, not two flavours of the same problem. A dependent pair (coincident lines, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ ) is always consistent — it has an entire line's worth of valid answers. An inconsistent pair (parallel lines, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$ ) has none at all. Algebraically the warning sign is identical at first glance — both cases make the variable vanish during elimination or substitution — but the statement left behind tells them apart instantly: a true statement with no variable (like $18=18$ ) means dependent and infinitely many solutions, while a false statement (like $0=9$ or $-4=0$ ) means inconsistent and no solution. Always read that final statement, not just the fact that the variable disappeared.

You should be able to…

State the general form of a linear equation in two variables and of a pair of such equations.
Explain, using a graph, what it means for a pair of linear equations to be consistent, inconsistent or dependent.
Match each of the three graphical cases — intersecting, parallel, coincident — to its correct count of solutions.
Apply the ratio test on $a_1/a_2$ , $b_1/b_2$ , $c_1/c_2$ to classify a pair of equations without graphing.
Solve a pair of linear equations by the substitution method, including recognising the true-statement and false-statement cases.
Solve a pair of linear equations by the elimination method, choosing a sensible variable and multiplier to eliminate first.
Translate a word problem — ages, digits, costs, incomes — into a pair of linear equations in two variables.
Verify a found solution against both original equations, and against every condition stated in a word problem.
Determine an unknown parameter (like $k$ ) that makes a given pair consistent, inconsistent or dependent.

Ask the tutor: Want to work through a fresh word problem together, forming the pair of equations from scratch and solving it by elimination?

How do I decide whether to use substitution or elimination in an exam?

Look at how the equations are already written. If one equation already has a variable with coefficient $1$ , or is easy to rearrange cleanly (like $x+2y=3$ ), substitution is usually quicker since the rearranging step is nearly free. If both equations have messier coefficients on both variables, elimination is usually faster, because you avoid ever writing an awkward fraction until the very last step. Neither method is more "correct" than the other — both always give the same answer for a consistent pair — so many students simply pick whichever leaves the smaller numbers to work with, and switch methods if the first choice starts getting messy.

Why does the chapter teach the graphical method first if it is not accurate enough to rely on?

The graphical method is taught first because it builds the mental picture every algebraic method depends on — that solving a pair of linear equations means finding where two lines meet, and that this can happen in exactly three ways. Once you can see intersecting, parallel and coincident lines clearly, the algebraic ratio test and the true or false statements that appear during substitution and elimination stop looking like arbitrary rules and start looking like exactly what you would expect from the picture. In practice, you graph only when a question explicitly asks you to, or when the numbers are small and clean; for anything else, including any answer that is not a tidy integer, an algebraic method is used instead.

What is the real difference between an inconsistent pair and a dependent pair of equations?

An inconsistent pair has absolutely no solution — its two lines are parallel and never meet, so no $(x,y)$ can satisfy both equations at once; algebraically, $a_1/a_2=b_1/b_2\neq c_1/c_2$ . A dependent pair, by contrast, has infinitely many solutions, because its two equations are secretly the same line written two different ways, so every single point on that line works; algebraically, $a_1/a_2=b_1/b_2=c_1/c_2$ . A dependent pair is always called consistent, since it does have solutions, just not a unique one — only the inconsistent case has zero solutions. When elimination makes the variable disappear, always check whether the leftover numeric statement is true or false to tell the two apart.

In the two-digit number problem, why are there two different correct answers?

The problem states that the sum of the number and its reverse is $66$ , giving one equation, $x+y=6$ after simplifying, and that the digits differ by $2$ , which is really two separate possibilities: either $x-y=2$ or $y-x=2$ , since the problem does not say which digit is larger. Solving the sum equation together with each of these in turn gives two different, equally valid pairs of linear equations, and hence two different numbers, $42$ and $24$ , both of which genuinely satisfy every condition stated in the problem. Whenever a word problem contains a phrase like "differ by," check if it is hiding two cases before writing down just one equation.