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Pair of Linear Equations in Two Variables - Algebraic Methods of Solving a Pair of Linear Equations

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A pair of linear equations in two variables xx and yy can be represented as a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0, where a1,b1,c1,a2,b2,c2a_1, b_1, c_1, a_2, b_2, c_2 are real numbers.

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The Substitution Method involves expressing one variable in terms of the other from one equation and substituting this value into the second equation to get a linear equation in one variable.

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The Elimination Method involves multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable (either xx or yy) become numerically equal. We then add or subtract the equations to eliminate that variable.

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A pair of equations is Consistent if it has at least one solution. It is Inconsistent if it has no solution.

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If the algebraic process results in a true statement like 0=00 = 0, the pair of equations has infinitely many solutions. If it results in a false statement like 0=50 = 5, the pair has no solution.

📐Formulae

a1x+b1y+c1=0a_1x + b_1y + c_1 = 0

a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

Unique Solution: a1a2≠b1b2\text{Unique Solution: } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}

Infinitely Many Solutions: a1a2=b1b2=c1c2\text{Infinitely Many Solutions: } \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

No Solution: a1a2=b1b2≠c1c2\text{No Solution: } \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

💡Examples

Problem 1:

Solve the following pair of equations using the substitution method: x+y=14x + y = 14 x−y=4x - y = 4

Solution:

From equation (2), we get x=4+yx = 4 + y. Substitute this value of xx in equation (1): (4+y)+y=14(4 + y) + y = 14 4+2y=144 + 2y = 14 2y=102y = 10 y=5y = 5. Now, substitute y=5y = 5 in x=4+yx = 4 + y: x=4+5=9x = 4 + 5 = 9. Therefore, the solution is x=9,y=5x = 9, y = 5.

Explanation:

We expressed xx in terms of yy using the simpler equation and substituted it into the other to reduce the system to one variable.

Problem 2:

Solve using the elimination method: 2x+3y=82x + 3y = 8 4x+5y=144x + 5y = 14

Solution:

To eliminate xx, multiply the first equation by 22: 2(2x+3y)=2(8)  ⟹  4x+6y=162(2x + 3y) = 2(8) \implies 4x + 6y = 16 (Equation 3). Subtract equation (2) from equation (3): 4x+6y=16−(4x+5y=14)y=2\begin{array}{r} 4x + 6y = 16 \\ -(4x + 5y = 14) \\ \hline y = 2 \end{array} Substitute y=2y = 2 in the first equation: 2x+3(2)=82x + 3(2) = 8 2x+6=82x + 6 = 8 2x=2  ⟹  x=12x = 2 \implies x = 1. Solution: x=1,y=2x = 1, y = 2.

Explanation:

We made the coefficients of xx equal in both equations by multiplying the first equation by 22, then subtracted them to find yy.