What is a graphical solution of linear equations? A pair of linear equations in two variables \(x\) and \(y\) represents two straight lines on the coordinate plane. The graphical solution is the point (or points) where these two lines meet. That meeting point gives the values of \(x\) and \(y\) that satisfy both equations at the same time.
General form. Any linear equation in two variables can be written as \(a_{1}x + b_{1}y + c_{1} = 0\) and \(a_{2}x + b_{2}y + c_{2} = 0\), where \(a_{1}, b_{1}\) are not both zero and \(a_{2}, b_{2}\) are not both zero. A solution of the pair is an ordered pair \((x, y)\) that makes both equations true.
Formulating a real situation. Many word problems reduce to a pair of lines. For example, "Aftab tells his daughter, seven years ago I was seven times as old as you were; three years from now I shall be three times as old as you will be." If \(x\) is Aftab's present age and \(y\) the daughter's, this becomes \(x - 7y + 42 = 0\) and \(x - 3y - 6 = 0\). Each condition is one straight line, and the common solution is their intersection.
How to draw a line. A straight line is fixed by any two points on it. Pick convenient \(x\)-values (often \(x = 0\) and \(y = 0\) give the intercepts), compute the matching \(y\), tabulate, plot and join with a ruler. Extend the line so the two graphs cross within the sheet.
Three possible cases. When you plot the two lines, exactly one of three things happens:
| Algebraic condition | Geometry | Solutions | Consistency |
|---|
| \(\dfrac{a_{1}}{a_{2}} \ne \dfrac{b_{1}}{b_{2}}\) | Intersecting lines | Exactly one (unique) | Consistent |
| \(\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}\) | Coincident lines | Infinitely many | Consistent (dependent) |
| \(\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \ne \dfrac{c_{1}}{c_{2}}\) | Parallel lines | None | Inconsistent |
Reading the ratio test. Always write each equation in the standard \(=0\) form first so that you have correct signs for \(c_{1}\) and \(c_{2}\). Then form the three ratios \(\dfrac{a_{1}}{a_{2}}\), \(\dfrac{b_{1}}{b_{2}}\), \(\dfrac{c_{1}}{c_{2}}\). If the first two differ, the lines cross once. If all three agree, the two equations are really the same line. If the coefficient ratios match but the constant ratio is different, the lines are parallel and never meet.
Consistent vs inconsistent. A system that has at least one solution is consistent (unique or infinitely many); a system with no solution is inconsistent. Coincident lines give a dependent consistent system, because every point on the line is a solution.
Real-life analogy. Think of two roads on a map. They either cross at one junction (unique solution), are literally the same road (every point common, infinitely many solutions), or run parallel and never meet (no solution).
- Cross once → unique solution, consistent.
- Same line → infinitely many solutions, dependent.
- Parallel → no solution, inconsistent.
Common mistakes to avoid. (1) Forgetting to move the constant to the left so \(c\) carries the right sign before taking ratios. (2) Plotting only one point per line — you need two. (3) Reading the intersection roughly and not verifying it in both original equations. (4) Confusing "coincident" (infinitely many) with "parallel" (none) — the difference is only in the constant ratio.
Solve graphically: \(x + y = 6\) and \(x - y = 2\).
Solution- Step 1: For \(x + y = 6\), take \(x=0 \Rightarrow y=6\) and \(y=0 \Rightarrow x=6\): points \((0,6)\) and \((6,0)\).
- Step 2: For \(x - y = 2\), take \(x=0 \Rightarrow y=-2\) and \(y=0 \Rightarrow x=2\): points \((0,-2)\) and \((2,0)\).
- Step 3: Plot both lines on graph paper and join each pair of points.
- Step 4: The lines meet at \((4,2)\).
- Step 5: Verify: \(4+2=6\) and \(4-2=2\), both true.
Answer: \(x=4,\ y=2\) (unique solution; lines intersect).
Without drawing, determine whether the pair \(2x + 3y - 6 = 0\) and \(4x + 6y - 12 = 0\) has a unique solution, no solution, or infinitely many.
Solution- Step 1: Identify coefficients: \(a_{1}=2, b_{1}=3, c_{1}=-6\); \(a_{2}=4, b_{2}=6, c_{2}=-12\).
- Step 2: Ratios: \(\dfrac{a_{1}}{a_{2}}=\dfrac{2}{4}=\dfrac{1}{2}\), \(\dfrac{b_{1}}{b_{2}}=\dfrac{3}{6}=\dfrac{1}{2}\), \(\dfrac{c_{1}}{c_{2}}=\dfrac{-6}{-12}=\dfrac{1}{2}\).
- Step 3: All three ratios are equal, so the lines are coincident.
Answer: Infinitely many solutions (consistent, dependent).
For what value of \(k\) will the lines \(3x + y = 5\) and \(6x + 2y = k\) be parallel (no solution)?
Solution- Step 1: Standard form: \(3x+y-5=0\) and \(6x+2y-k=0\). So \(a_{1}=3,b_{1}=1,c_{1}=-5\); \(a_{2}=6,b_{2}=2,c_{2}=-k\).
- Step 2: For parallel lines need \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\ne\dfrac{c_{1}}{c_{2}}\).
- Step 3: \(\dfrac{3}{6}=\dfrac12\) and \(\dfrac{1}{2}=\dfrac12\), so the coefficient ratios match.
- Step 4: Need \(\dfrac{c_{1}}{c_{2}}=\dfrac{-5}{-k}=\dfrac{5}{k}\ne\dfrac12\), i.e. \(k\ne 10\).
Answer: \(k \ne 10\) (any value except \(10\)).
Check graphically whether \(x - 2y = 0\) and \(3x + 4y = 20\) are consistent, and find the solution.
Solution- Step 1: For \(x-2y=0\): \(x=0\Rightarrow y=0\); \(x=4\Rightarrow y=2\). Points \((0,0)\) and \((4,2)\).
- Step 2: For \(3x+4y=20\): \(x=0\Rightarrow y=5\); \(y=0\Rightarrow x=\tfrac{20}{3}\approx 6.7\). Use also \(x=4\Rightarrow y=2\).
- Step 3: Both lines pass through \((4,2)\), so they intersect there.
- Step 4: Verify: \(4-2(2)=0\) and \(3(4)+4(2)=12+8=20\), both true.
Answer: Consistent; unique solution \(x=4,\ y=2\).
Show that \(x + y = 5\) and \(2x + 2y = 10\) represent the same line.
Solution- Step 1: Standard form: \(x+y-5=0\), \(2x+2y-10=0\).
- Step 2: Ratios: \(\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{-5}{-10}=\dfrac12\).
- Step 3: All ratios equal, so the second equation is just twice the first.
- Step 4: Every point of \(x+y=5\) satisfies both, e.g. \((2,3)\) and \((5,0)\).
Answer: Coincident lines; infinitely many solutions.
Half the perimeter of a rectangular garden is \(36\) m. The length is \(4\) m more than the width. Form a pair of linear equations and find the dimensions graphically.
Solution- Step 1: Let length \(=x\) m, width \(=y\) m.
- Step 2: Half-perimeter: \(x+y=36\). Length condition: \(x=y+4\), i.e. \(x-y=4\).
- Step 3: For \(x+y=36\): \((0,36),(36,0)\). For \(x-y=4\): \((4,0),(0,-4)\).
- Step 4: Lines meet at \((20,16)\); verify \(20+16=36\) and \(20-16=4\).
Answer: Length \(=20\) m, width \(=16\) m.
Determine, by comparing ratios, the nature of the pair \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\).
Solution- Step 1: \(a_{1}=5,b_{1}=-4,c_{1}=8\); \(a_{2}=7,b_{2}=6,c_{2}=-9\).
- Step 2: \(\dfrac{a_{1}}{a_{2}}=\dfrac{5}{7}\), \(\dfrac{b_{1}}{b_{2}}=\dfrac{-4}{6}=-\dfrac{2}{3}\).
- Step 3: Since \(\dfrac{5}{7}\ne -\dfrac{2}{3}\), the coefficient ratios differ.
- Step 4: Unequal coefficient ratios mean the lines intersect at exactly one point.
Answer: Intersecting lines → unique solution (consistent).
For what value of \(k\) do \(2x + 3y = 7\) and \((k-1)x + (k+2)y = 3k\) have infinitely many solutions?
Solution- Step 1: Condition: \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}\), i.e. \(\dfrac{2}{k-1}=\dfrac{3}{k+2}=\dfrac{7}{3k}\).
- Step 2: From \(\dfrac{2}{k-1}=\dfrac{3}{k+2}\): \(2(k+2)=3(k-1)\Rightarrow 2k+4=3k-3\Rightarrow k=7\).
- Step 3: Check with the third ratio: \(\dfrac{2}{6}=\dfrac13\) and \(\dfrac{7}{21}=\dfrac13\), they agree.
Answer: \(k=7\).
A pair of lines is \(x + 2y = 5\) and \(3x + 12y = 10\). Are they parallel? Decide using ratios.
Solution- Step 1: \(a_{1}=1,b_{1}=2,c_{1}=-5\); \(a_{2}=3,b_{2}=12,c_{2}=-10\).
- Step 2: \(\dfrac{a_{1}}{a_{2}}=\dfrac{1}{3}\), \(\dfrac{b_{1}}{b_{2}}=\dfrac{2}{12}=\dfrac16\).
- Step 3: \(\dfrac13\ne\dfrac16\), so the coefficient ratios are unequal.
- Step 4: Unequal ratios ⇒ not parallel; the lines intersect once.
Answer: Not parallel; they intersect (unique solution).
Solve graphically \(x + y = 7\) and \(3x - y = 1\), and state the coordinates of the intersection.
Solution- Step 1: For \(x+y=7\): \((0,7),(7,0)\).
- Step 2: For \(3x-y=1\): \(x=0\Rightarrow y=-1\); \(x=1\Rightarrow y=2\). Points \((0,-1),(1,2)\).
- Step 3: Plot and join; the lines cross at \((2,5)\).
- Step 4: Verify: \(2+5=7\) and \(3(2)-5=1\), both true.
Answer: \((x,y)=(2,5)\).
The coach buys \(3\) bats and \(6\) balls for ₹3900; later \(1\) bat and \(2\) balls for ₹1300. Set up the pair, decide its nature by ratios, and explain.
Solution- Step 1: Let bat \(=x\), ball \(=y\) (in rupees). Equations: \(3x+6y=3900\) and \(x+2y=1300\).
- Step 2: Standard form: \(3x+6y-3900=0\), \(x+2y-1300=0\).
- Step 3: Ratios: \(\dfrac{3}{1}=3,\ \dfrac{6}{2}=3,\ \dfrac{-3900}{-1300}=3\).
- Step 4: All three equal ⇒ coincident lines, so the two purchases give the same relation.
Answer: Infinitely many solutions; the data is dependent (one relation, not two independent prices).
Find the value(s) of \(k\) for which \(kx + 2y = 5\) and \(3x + y = 1\) have a unique solution.
Solution- Step 1: \(a_{1}=k,b_{1}=2\); \(a_{2}=3,b_{2}=1\).
- Step 2: Unique solution requires \(\dfrac{a_{1}}{a_{2}}\ne\dfrac{b_{1}}{b_{2}}\), i.e. \(\dfrac{k}{3}\ne\dfrac{2}{1}\).
- Step 3: So \(k\ne 6\).
Answer: \(k \ne 6\) (any value except \(6\) gives a unique solution).