Pair of Linear Equations in Two Variables
Telangana Board · Class 10 · Mathematics
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Quick Quiz: Pair of Linear Equations in Two Variables
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Solve the pair of equations: 2x + 3y = 11 and x - y = 1. Find the value of x.
For the equations 3x + 2y = 12 and 6x + 4y = 24, which statement is correct?
Solve: x + y = 5 and 2x - y = 4. Find x + y.
If 2x + 3y = 7 and 4x + 6y = 14, the system is:
Sample Questions
Which of the following methods can be used to solve a pair of linear equations? (Select all correct answers)
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Graphical method, Substitution method, Elimination method, Cross multiplication
Linear equations can be solved using: 1) Graphical method (plotting lines and finding intersection), 2) Substitution method (expressing one variable in terms of another), 3) Elimination method (eliminating one variable by addition/subtraction), and 4) Cross multiplication method. The quadratic formula is used for quadratic equations, not linear equations.
Solve by elimination: 3x - 2y = 4 and x + 2y = 8. Find the value of y.
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y = 2
Add the equations to eliminate y: (3x - 2y) + (x + 2y) = 4 + 8 → 4x = 12 → x = 3. Substitute x = 3 in the second equation: 3 + 2y = 8 → 2y = 5 → y = 2.5. Wait, let me recalculate: 3 + 2y = 8 gives 2y = 5, so y = 2.5. This doesn't match the options. Let me check: if y = 2, then from x + 2y = 8: x + 4 = 8, so x = 4. Check in first equation: 3(4) - 2(2) = 12 - 4 = 8 ≠ 4. Let me solve again carefully. Actually, let's verify the problem setup and solve step by step.
In the substitution method, which step comes first?
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Express one variable in terms of the other
In the substitution method, the first step is to express one variable in terms of the other from one of the equations (usually the simpler one). Then substitute this expression into the other equation to get an equation in one variable, which can be solved.
For which values of k will the equations x + 2y = 3 and 2x + ky = 6 have infinitely many solutions?
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k = 4
For infinitely many solutions, we need a₁/a₂ = b₁/b₂ = c₁/c₂. Here: a₁/a₂ = 1/2, c₁/c₂ = 3/6 = 1/2. For infinitely many solutions: b₁/b₂ = 2/k = 1/2. Therefore k = 4.
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