NCERT Solutions Class 10 Maths Chapter 3 Linear Equations

Translate each sentence into one equation, then let the graph confirm the meeting point. The first job in both parts is to name the two unknowns and write exactly two equations, because a graphical solution needs a pair of lines and nothing more.

• Word to equation: keep the wording in the same order it is written. In part (i) "girls are 4 more than boys" becomes y=x+4 and "ten students in all" becomes x+y=10, which saves a student from the common sign slip of writing the difference the wrong way round as y-x=-4.
• Read the crossing: plot each line through two neat points and read where they cross. In part (i) the lines meet at (3,7), giving 3 boys and 7 girls. In part (ii) the two cost lines both pass through (3,5), so the pencil is Rs. 3 and the pen is Rs. 5.
• Where the marks sit: the examiner rewards three separate things, the correctly formed pair of equations, a small table of two points for each line, and a clearly marked meeting point, so a student should label that crossing on the graph itself and not merely state it in words below the diagram where it is easy to miss.
• Defend the answer: put the read-off point back into both of the original equations as a final step. A graph that is drawn slightly off the true crossing can still be defended by that one line of checking, since here the totals 3+7=10 and the difference 7-3=4 both come out exactly right and confirm the reading.

Boys 3, girls 7; pencil Rs. 3, pen Rs. 5.

Compare the ratios in a fixed order and stop at the first that differs. The whole question is one rule applied three times, so the work is fast once the order of checking is fixed.

• Fixed order: write all three lines in the form ax+by+c=0, then compute the first two ratios a₁/a₂ and b₁/b₂. If those two differ, as in part (i), the lines intersect and there is no need to look at the constant ratio at all.
• Then the constant: only when the first two match do you check the third ratio. Equal everywhere means coincident as in part (ii), while the constant breaking the pattern means parallel as in part (iii).
• Sign trap: the most common error is dropping a minus, because b₁=-4 and b₂=-1 must keep their signs, and rounding the constant ratio 109 in part (iii) to a careless 1 would wrongly turn parallel lines into coincident ones.
• Earning the marks: show each ratio reduced to lowest terms and then write the one word conclusion, since the reasoning marks sit in the comparison rather than in the verdict on its own.

Intersecting; coincident; parallel.

Consistent or inconsistent is decided by the ratio test, so classify, never solve. This question never asks for the actual values of x and y, only whether a solution exists, so a student should resist solving and read off the three ratios instead.

• Clean routine: write each equation as ax+by+c=0, simplify any fractions, and compare a₁/a₂ with b₁/b₂ before touching the constant ratio.
• The verdicts: parts (i) and (iii) have unequal first two ratios, so they are consistent with a unique solution; part (ii) matches in the first two but not the third, the exact signature of an inconsistent pair; parts (iv) and (v) match in all three, so they are consistent with infinitely many solutions.
• Fraction care: the fractional coefficients in (iii) and (v) are where errors creep in, so dividing 32 by 9 to get 16 should be done carefully, and in (iv) keeping the minus on -10 and -22 is what makes all three ratios equal.
• Full marks: state the three ratios, the comparison and the one word verdict, and add whether a consistent pair has one solution or infinitely many, since that distinction often carries a mark of its own.

Consistent: (i), (iii), (iv), (v). Inconsistent: (ii).

Classify with the ratios, then graph only the one part that needs it. The smart order here is to apply the ratio test to all four pairs first, because three of them are settled without a single plotted point.

• Settle by ratio: part (i) has all three ratios equal, so the lines coincide and the pair is consistent with infinitely many solutions; parts (ii) and (iv) match in the first two ratios but not the third, the fingerprint of parallel lines, so both are inconsistent.
• Graph only one: only part (iii) has unequal first two ratios, which guarantees a single crossing point, so that is the only pair worth drawing, and plotting two clean points on each line gives the meeting at (2,2).
• Check the point: put that point into both equations, since 2(2)+2=6 and 4(2)-2(2)=4 are both true, which confirms the read-off before any marks are claimed.
• Standard form first: in parts (ii) and (iv), write each equation as ax+by+c=0 before reading the constant, because a term left on the wrong side flips the third ratio and could turn a parallel pair into a coincident one. Simplifying also helps, since dividing 2x-2y-2=0 by 2 gives x-y-1=0 and makes the comparison obvious.
• Marks rule: state every verdict with its ratio reasoning, and supply a graph only where a unique solution exists, because graphing the inconsistent pairs earns nothing while skipping the one unique pair loses the plotting marks.

Consistent: (i) infinite, (iii) x=2,y=2. Inconsistent: (ii), (iv).

Read "half the perimeter" as length plus width to avoid an extra step. The one idea that simplifies this problem is recognising that half of 2(l+w) is just l+w.

• Clean first equation: half the perimeter gives the tidy x+y=36 straight away, rather than the longer 2(x+y)=72 that you would then have to divide down.
• Add, do not substitute: the second condition gives x-y=4, and adding the two equations cancels y at once to give x=20, after which y=16 follows. Adding is faster here because the y terms are already +y and -y, so they vanish with no multiplication.
• Protect the marks: label which letter is length and which is width, state both equations, show the elimination, and finish with a check that 20+16=36 and 20-16=4. Forgetting both dimensions or mixing up length and width are the two slips that cost marks, so naming the variables clearly at the start guards the answer.

Length =20 m, width =16 m.

Engineer the second equation to fit the ratio rule for each case. The question is open ended, so the goal is not a unique answer but a second equation that provably meets each geometric condition, and the cleanest way to guarantee that is to work backwards from the ratio test.

• Intersecting: any choice with unequal first two ratios works, so simply swapping the coefficients to give 3x+2y-7=0 is a safe pick that is easy to justify.
• Parallel: keep a and b in the same ratio as the original, easiest by multiplying 2x+3y by a constant, and then choose a constant term that breaks the third ratio, which 4x+6y-9=0 does.
• Coincident: the surest move is to multiply the entire given equation by a non-zero number, here 2, giving 4x+6y-16=0, because every ratio then equals that same number automatically.
• Show the why: state the ratio condition being targeted before writing each equation and confirm the ratios afterwards, since the marks are for the justification rather than the particular numbers chosen, and an examiner accepts any equation that satisfies the stated condition with working shown.

Intersecting 3x+2y-7=0; parallel 4x+6y-9=0; coincident 4x+6y-16=0.

Find the three corners as three separate intersections, then shade. The triangle here is bounded by three lines, the two given equations and the x-axis, so its three vertices are exactly the three pairwise crossings.

• The apex: the top corner is where the two given lines meet, and solving them together gives (2,3), the single point that satisfies both equations at once.
• The base corners: the base lies along the x-axis, so the other two vertices come from setting y=0 in each equation, which gives (-1,0) from the first line and (4,0) from the second.
• Draw and shade: a student who draws both lines accurately through two plotted points each will see the same three corners, and shading the enclosed region answers the final part of the question.
• Where marks live: the graph must show both lines labelled, the three vertices marked with their coordinates, and the region clearly shaded, because stating the vertices without the diagram, or the diagram without the coordinates, leaves easy marks on the table.
• Sanity check: the base from x=-1 to x=4 has length 5 and the height to the apex is 3, so the area is 1253=7.5 square units, a tidy confirmation that the three corners are right.

Vertices (2,3), (-1,0), (4,0).

Tidy the coefficients before substituting, and read a vanishing variable correctly. Substitution is reliable across all six parts, but the practical skill is preparing each pair first.

• Clear the mess: multiply by 10 to remove the decimals in part (iv), and by 6 to remove the fractions in parts (ii) and (vi), so the substitution then runs over whole numbers and the chance of an arithmetic slip drops sharply.
• Read an identity: part (iii) collapses to 9=9, true for every value, which means the two equations are the same line and there are infinitely many solutions, not "no answer". A student must state that conclusion rather than leave the problem blank, because the mark is for recognising the dependent pair.
• The zero case: part (v) is the mirror, since both constants are zero, so the lines pass through the origin and substitution forces y=0 and then x=0, the single solution (0,0). The surds look frightening but never need to be evaluated, because the combined coefficient is simply non-zero.
• Isolate wisely: pick the equation where a coefficient is one or easy, as with x=4+y in part (i) and s=3+t in part (ii), since that keeps the algebra clean and the substitution short.
• Board habit: clear the fractions on a visible line, isolate one variable, substitute, solve, and back-substitute, finishing each part with both values stated together and a one-line check that puts the pair back into the original equations.

x=9,y=5; s=9,t=6; infinite; x=2,y=3; x=0,y=0; x=2,y=3.

Get the solution point first, because m is decided only after x and y are known. The question has two stages, and the order matters: solve the pair, then use the point to find m.

• Eliminate x: since both equations carry 2x, subtracting one from the other removes x in a single step and gives y=5, after which x=-2 follows from either original equation.
• Fit the slope: the relation y=mx+3 is just one more condition the same point must obey, so substituting x=-2 and y=5 turns it into 5=-2m+3, which solves to m=-1.
• Avoid the trap: a frequent error is to chase m before solving for the point, or to drop the values into the wrong slots, so writing the relation with y=5 and x=-2 clearly labelled prevents that, and the check 5=(-1)(-2)+3=2+3=5 confirms the slope. Present the elimination, the solved point, and the substitution for m as three clear stages.

x=-2, y=5, m=-1.

Translate each clue into an equation in the order it is written, then substitute the simplest variable. Every part of this question rewards careful reading far more than heavy algebra, so the time is best spent on the set-up.

• Isolate what is given: "difference is 26" and "one is three times the other" give x-y=26 and x=3y, and because x is already isolated in the second relation, the substitution into the first is instant and needs no rearranging.
• Recall the fact: the supplementary angles in part (ii) hinge on the angles summing to 180^∘, the one piece of recalled knowledge the question tests, and a student who confuses supplementary with complementary uses 90^∘ and gets both angles wrong.
• Tame the big numbers: the bats and balls pair in part (iii) has no modelling trick, but the numbers are large, so isolating x from the simpler equation 3x+5y=1750 keeps the substitution manageable and avoids unwieldy fractions partway through.
• Rearrange first: the fraction problem (v) and the age problem (vi) each need a small rearrangement, cross-multiplying x+2y+2=911 into 11x-9y=-4, and turning "three times" and "seven times" into linear forms, after which the substitution is routine.
• Shift both ages: a common slip in age problems is to apply the "five years ago" and "five years hence" change to only one person, so each age must shift by the same amount before the multiple is taken.
• Answer everything: the taxi problem (iv) has a final part students often skip, the charge for 25 km, so after finding the fixed charge 5 and rate 10 the answer must include 5+2510=255. Reading the question to the very end is what secures the last mark in parts (iv) and (vi).
• Verify each part: name both unknowns clearly, write the two equations, substitute, solve, and finish each part with a quick check such as 7+2 over 9+2 giving 911, since one verifying line per part catches an arithmetic slip before it ever costs the accuracy mark on that part of the answer.

39,13; 99^∘,81^∘; bat 500, ball 50; fixed 5, rate 10, 25 km =255; 79; Jacob 40, son 10.

Pick the variable that is cheapest to eliminate, and always tidy fractions first. Both methods are required by the question, but elimination is usually the quicker route when the coefficients can be matched with a small multiplier, and the real skill is spotting that match before any arithmetic.

• Find the match: in part (i), tripling the first equation lines up the terms in y; in part (ii), doubling the second matches the term in y; in part (iii), tripling the first matches the term in x, so the right multiplier removes a variable in one clean step every time.
• Clear before eliminating: in part (iv), clearing the fractions first to 3x+4y=-6 and 3x-y=9 makes the terms in x cancel directly by subtraction, which is far tidier than wrestling with fractional coefficients while you eliminate.
• Mind the signs: keep signs straight when subtracting, since -13y=5 in part (iii) gives a negative value, not a positive one, and a single sign slip there cascades into a wrong value for the other variable as well.
• Check and present: after finding both values, substituting them back into either original equation is a fast check, such as 3(2)+4(1)=10 in part (ii). The marks reward a clear elimination step, the solved variable, the back-substitution and the final pair, so write each stage on its own line to keep the method transparent.

195,65; 2,1; 913,-513; 2,-3.

Name the unknowns to match the question, then simplify before eliminating. The hard part of each problem is the translation, not the arithmetic, so the unknowns should be chosen to fit exactly what the question asks.

• Fraction care: in part (i) the two conditions become x-y=-2 and 2x-y=1, and a student must take care that "denominator subtract 1" lands on the denominator, not the numerator, since swapping the two changes the whole answer.
• Shift both ages: in part (ii) each person's age must shift by the same five or ten years before the multiple "thrice" or "twice" is applied, so the relation is x-5=3(y-5) rather than x-5=3y, and getting that shift onto both ages is the most common stumbling point.
• Use place value: in part (iii) the number is 10x+y and its reverse 10y+x, and writing them this way turns "nine times the number is twice the reverse" into 88x-11y=0, which divides down to the tidy 8x-y=0. Treating the number as just x+y is the error that wrecks this part.
• Simplify and read clauses: parts (iv) and (v) both simplify before elimination, since 50x+100y=2000 reduces to x+2y=40, while "seven days" means four extra days beyond the first three and "five days" means two extra, not the full counts.
• Eliminate then check: once each pair is in clean form, subtracting to remove the matching variable is one quick step, and back-substitution gives the second value. Verify it, for instance that ten fifties and fifteen hundreds total 25 notes and Rs. 2000, or that the digits of 18 add to 9 and nine times 18 equals twice 81.
• The recurring trap: across all five parts the real danger is mis-modelling the words, especially the "first three days" clause, the equal shift in ages, and the place value of a two-digit number, so a careful set-up wins the marks and a quick verification secures them.

35; Nuri 50, Sonu 20; 18; ten Rs. 50 and fifteen Rs. 100; fixed 15, per extra day 3.