Pair of Linear Equations Class 10 Exercise 3.1 Solutions

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).