Pair of Linear Equations in Two Variables
This chapter covers pairs of linear equations in two variables, solving them graphically and algebraically (substitution and elimination methods), consistency conditions, and word problems.
Two linear equations of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 (with a1, a2, b1, b2 not both zero in each) together form a pair of linear equations in two variables. Each represents a straight line, and their common solution is the point where the two lines meet.
For a1x+b1y+c1=0 and a2x+b2y+c2=0: if a1/a2 ≠ b1/b2, the lines intersect at a unique point (exactly one solution, consistent). If a1/a2 = b1/b2 = c1/c2, the lines are coincident (infinitely many solutions, consistent). If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel (no solution, inconsistent).
Check graphically whether the pair of equations x + 3y = 6 and 2x - 3y = 12 is consistent. If so, solve them graphically.
a₁/a₂ = 1/2, b₁/b₂ = 3/-3 = -1. Since 1/2 ≠ -1, the lines intersect at a unique point - the pair is consistent. Plotting: x+3y=6 passes through (0,2) and (6,0); 2x-3y=12 passes through (0,-4) and (6,0). They intersect at (6, 0).
On comparing the ratios a₁/a₂, b₁/b₂, c₁/c₂, find whether the lines 5x-4y+8=0 and 7x+6y-9=0 are consistent, inconsistent, or dependent.
a₁/a₂ = 5/7, b₁/b₂ = -4/6 = -2/3. Since 5/7 ≠ -2/3, the lines intersect at a unique point - consistent with a unique solution.
Solve the pair of linear equations by the substitution method: x + y = 14, x - y = 4.
From the first equation: x = 14 - y. Substitute into the second: (14-y) - y = 4, so 14 - 2y = 4, giving 2y = 10, y = 5. Then x = 14 - 5 = 9. Solution: x = 9, y = 5.
Solve the pair of linear equations by the substitution method: s - t = 3, s/3 + t/2 = 6.
From the first equation: s = t + 3. Substitute: (t+3)/3 + t/2 = 6. Multiply through by 6: 2(t+3) + 3t = 36, so 2t+6+3t=36, 5t=30, t=6. Then s = 6+3 = 9. Solution: s = 9, t = 6.
Solve by substitution: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3.
Multiply both by 10: 2x+3y=13 and 4x+5y=23. From first: x = (13-3y)/2. Substitute: 4((13-3y)/2) + 5y = 23, so 2(13-3y)+5y=23, 26-6y+5y=23, 26-y=23, y=3. Then x = (13-9)/2 = 2. Solution: x = 2, y = 3.
Solve by the elimination method: x + y = 5 and 2x - 3y = 4.
Multiply the first equation by 3: 3x+3y=15. Add to the second: (2x-3y)+(3x+3y) = 4+15, so 5x = 19, x = 19/5. Then from x+y=5: y = 5 - 19/5 = 6/5. Solution: x = 19/5, y = 6/5.
Solve by the elimination method: 3x + 4y = 10 and 2x - 2y = 2.
Multiply the second equation by 2: 4x-4y=4. Add to the first: (3x+4y)+(4x-4y)=10+4, so 7x=14, x=2. Substitute into 2x-2y=2: 4-2y=2, 2y=2, y=1. Solution: x = 2, y = 1.
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let the digits be x (tens) and y (units), so the number is 10x+y. x+y=9. Also 9(10x+y) = 2(10y+x), so 90x+9y=20y+2x, giving 88x=11y, so y=8x. Substituting into x+y=9: x+8x=9, 9x=9, x=1, y=8. The number is 18.
The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4:3. If each saves Rs 2000 per month, find their monthly incomes.
Let incomes be 9x and 7x, expenditures 4y and 3y. 9x - 4y = 2000 and 7x - 3y = 2000. From these: multiply first by 3, second by 4: 27x-12y=6000, 28x-12y=8000. Subtracting: x=2000. Then 9(2000)-4y=2000, 18000-4y=2000, 4y=16000, y=4000. Incomes: 9x = Rs 18000 and 7x = Rs 14000.
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Let speeds be x and y km/h (x > y), from A and B respectively. Same direction: 5x - 5y = 100, so x - y = 20. Towards each other: x + y = 100. Adding: 2x = 120, x = 60. Then y = 40. Speeds: 60 km/h and 40 km/h.
A pair of linear equations can be solved graphically (plotting both lines and reading the intersection), or algebraically using the substitution method or the elimination method. Algebraic methods give exact answers, while graphical solutions can only be read approximately.
If a pair of linear equations is consistent, then the lines represented by them are:
Correct answer: (C) either intersecting or coincident.
The pair of equations x = 0 and y = 0 has:
Correct answer: (A) a unique solution, (0, 0) - the point where the y-axis and x-axis meet.
If a pair of linear equations represents parallel lines, the pair has no solution.
True - parallel lines never intersect, so there is no common point satisfying both equations.
If a₁/a₂ = b₁/b₂ = c₁/c₂ for a pair of linear equations, the lines are _______ and the pair has _______ solutions.
coincident; infinitely many