CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 11

Linear Equation in Two Variables

An equation of the form Ax + By + C = 0, where A, B, C are real numbers and A and B are not both zero, is a linear equation in two variables. Both unknowns appear only to the first power and are never multiplied together.
If C = 0 the line passes through the origin. If B = 0 the equation reduces to Ax + C = 0 (a vertical line) and if A = 0 it reduces to By + C = 0 (a horizontal line).
One equation in two unknowns has infinitely many solutions, so a pair of equations is needed to fix a single pair of values.

Methods of Solving a Pair

For a pair \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) there are three standard methods.
1. Substitution: express one variable from one equation and put it in the other.
2. Elimination: multiply the equations by suitable numbers so that one variable has equal coefficients, then add or subtract.
3. Cross multiplication: \( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \), valid when \( a_1b_2 - a_2b_1 \neq 0 \).

Conditions for Solvability

1. Unique solution (intersecting lines, consistent): \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
2. No solution (parallel lines, inconsistent): \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
3. Infinitely many solutions (coincident lines, consistent): \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
To find an unknown constant in a question, write the required condition and solve the resulting equation, then check the remaining ratio.

Reducible Equations

Equations in which x and y occur in denominators, for example \( \frac{6}{x+y} = \frac{7}{x-y} + 3 \), are made linear by substituting new variables such as \( u = \frac{1}{x+y} \) and \( v = \frac{1}{x-y} \), or \( a = x + y \) and \( b = x - y \). Solve for the new variables and then return to x and y. Remember the restrictions such as x + y ≠ 0.

Word Problems

Choose two unknowns, translate each statement into an equation, solve the pair and check the answer against the story.
Two-digit numbers: write the number as 10x + y; the reversed number is 10y + x.
Ages: use present ages and write one equation for each time reference (past or future).
Fractions: fix numerator and denominator as the two unknowns.

Solved Examples

Example 1. Solve x + 4y = 14 and 7x − 3y = 5.
Answer:
From the first equation x = 14 − 4y.
Substitute: 7(14 − 4y) − 3y = 5, so 98 − 31y = 5 and y = 3.
Then x = 14 − 12 = 2. x = 2, y = 3

Example 2. Solve 9x − 4y = 8 and 13x + 7y = 101 by elimination.
Answer:
Multiply the first by 7 and the second by 4: 63x − 28y = 56 and 52x + 28y = 404.
Add: 115x = 460, so x = 4. Then 36 − 4y = 8 gives y = 7.
x = 4, y = 7

Example 3. Solve 3x + 2y + 25 = 0 and x + y + 15 = 0 by cross multiplication.
Answer:
a₁ = 3, b₁ = 2, c₁ = 25 and a₂ = 1, b₂ = 1, c₂ = 15.
\( \frac{x}{2 \times 15 - 1 \times 25} = \frac{y}{25 \times 1 - 15 \times 3} = \frac{1}{3 \times 1 - 2 \times 1} \), that is \( \frac{x}{5} = \frac{y}{-20} = 1 \).
x = 5, y = −20

Example 4. Solve \( \frac{6}{x+y} = \frac{7}{x-y} + 3 \) and \( \frac{1}{2(x+y)} = \frac{1}{3(x-y)} \), where x + y ≠ 0 and x − y ≠ 0.
Answer:
Put a = x + y and b = x − y. The second equation gives 3b = 2a, so a = 3b/2.
The first becomes 6/a = 7/b + 3. With a = 3b/2: 4/b = 7/b + 3, so −3/b = 3 and b = −1. Then a = −3/2.
x = (a + b)/2 = −5/4 and y = (a − b)/2 = −1/4. x = −5/4, y = −1/4

Example 5. If \( \frac{1}{2(x+2y)} + \frac{5}{3(3x-2y)} = -\frac{3}{2} \) and \( \frac{5}{4(x+2y)} - \frac{3}{5(3x-2y)} = \frac{61}{60} \), find x and y.
Answer:
Let U = 1/(x + 2y) and V = 1/(3x − 2y). Then 3U + 10V = −9 and 75U − 36V = 61.
Multiply the first by 25 and subtract: 286V = −286, so V = −1 and U = 1/3.
x + 2y = 3 and 3x − 2y = −1. Adding: 4x = 2, so x = 1/2 and y = 5/4. x = 1/2, y = 5/4

Example 6. Find the values of M for which Mx − 2y = 9 and 4x − y = 7 have a unique solution.
Answer:
Unique solution needs M/4 ≠ (−2)/(−1) = 2.
So M ≠ 8. M can be any real number except 8.

Example 7. For what value of a does the system ax + 3y = a − 3 and 12x + ay = a have no solution?
Answer:
No solution needs a/12 = 3/a ≠ (a − 3)/a.
a² = 36, so a = 6 or −6. For a = 6 all three ratios are equal (infinite solutions), so it is rejected.
a = −6

Example 8. Find α and β for which 2x + 3y = 7 and 2αx + (α + β)y = 28 have infinitely many solutions.
Answer:
2/(2α) = 3/(α + β) = 7/28 = 1/4.
From 1/α = 1/4: α = 4. From 3/(α + β) = 1/4: α + β = 12, so β = 8.
α = 4, β = 8

Example 9. A two-digit number has the product of its digits equal to 18. When 63 is subtracted from the number, the digits interchange. Find the number.
Answer:
Let the number be 10x + y with xy = 18. Then 10x + y − 63 = 10y + x, so x − y = 7.
x = y + 7 gives y(y + 7) = 18, so y² + 7y − 18 = 0 and y = 2 (y = −9 is rejected).
x = 9. The number is 92.

Example 10. The sum of two numbers is 2490. If 6.5% of one number equals 8.5% of the other, find the numbers.
Answer:
x + y = 2490 and 6.5x = 8.5y, so x = (17/13)y.
(17/13)y + y = 2490, so y = 13 × 2490/30 = 1079 and x = 1411.
The numbers are 1411 and 1079.

Example 11. A says to B: if you give me 30 of your mangoes I will have twice as many as you are left with. B replies: if you give me 10, I will have three times as many as you are left with. How many mangoes does each have?
Answer:
Let A have x and B have y. Then x + 30 = 2(y − 30), so x − 2y = −90, and 3(x − 10) = y + 10 gives 3x − y = 40.
From the second, y = 3x − 40. Then x − 6x + 80 = −90, so x = 34 and y = 62.
A has 34 mangoes and B has 62 mangoes.