Class 10 Mathematics Chapter 3 Pair Of Linear Equation In Two Variable

Key Concept: Coefficient Matching

b) 2

Solution Description: To eliminate the variable, adjust the second equation so that the coefficients of x in both equations match. The first equation has a coefficient of 4 for x, while the second equation has a coefficient of 2. To make them equal, multiply the entire second equation by 2. This ensures that both equations have the same coefficient for x, allowing for elimination by subtraction.

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Key Concepts: Unique Solution

a) Yes

Solution Description:

To determine if the given system has a unique solution, we analyze the slopes of the two lines represented by the equations. If the lines have different slopes, they intersect at a single point, meaning the system has a unique solution.

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Key Concept: Solution Verification

a) Yes

Solution Description:

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Key Concepts: Consistent Definition

c) x + y = 5 and 2x + 2y = 10

Solution Description:

A system of linear equations is considered consistent if it has at least one solution, meaning the equations represent lines that intersect at a single point or overlap entirely. To determine consistency, compare the slopes and intercepts of the equations.

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Key Concept: Elimination Step

(b) Subtract the second equation from the first

Solution Description: To eliminate the second variable, multiply both sides of the first equation by a specific number and the second equation by another number so that the coefficients of the chosen variable become equal. Here, multiplying the first equation by a certain factor and the second by another factor ensures that the coefficients of the second variable are the same.

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Key Concept: Intersecting Lines

c) 2x + 3y = 9 and 4x + 6y = 18

Solution Description:

To determine if two lines intersect, compare their slopes. If the slopes are different, the lines intersect at a single point. If the slopes are the same but the intercepts differ, they are parallel and never meet. If both the slopes and intercepts are identical, the lines overlap completely.

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Key Concept: Simple Substitution

b) x = 6 − 3 2 y

Solution Description: To express one variable in terms of another, we begin by isolating the desired variable. Here, we aim to express x in terms of y. We start by moving all terms involving y to the opposite side of the equation. This ensures that x remains alone on one side. Next, we eliminate any coefficients attached to x by performing the necessary division. This step ensures that x is explicitly written in terms of y.

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Key Concept: Equation Form

b) $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$

Solution Description: A pair of linear equations in two variables represents two straight lines on a coordinate plane. The general form expresses each equation as a relationship between two variables. The solution to these equations is the point where the two lines intersect. There are three possible outcomes: a unique solution if the lines intersect at a single point, infinitely many solutions if the lines coincide, and no solution if the lines are parallel but distinct.

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Key Concept: Real-World Application

d) 30

Solution Description: The farmer sells two types of fruits: apples and bananas, each at different prices. On a particular day, he sold a total of 50 fruits and earned \$120 from the sales. To determine the number of bananas sold, we analyze the relationship between the total number of items and the total revenue. Since apples are more expensive than bananas, we need to find a combination where the sum of apples and bananas equals 50, and their total sales amount matches \$120. By systematically testing possibilities or using logical deduction, we find that selling more bananas and fewer apples balances the total earnings correctly.

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Key Concept: Complex Elimination

c) There are infinitely many solutions

[Solution Description] Multiply the first equation by 2 to make the coefficients of x equal: $6x + 8y = 24$. Subtracting the second equation from the result gives 0 = 0, a true statement with no variables, implying infinitely many solutions.

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Key Concepts: Coefficient Analysis

c) Infinitely many solutions

Solution Description:

The given equations have proportional coefficients, meaning one equation is a multiple of the other. This indicates that both represent the same line. When two equations describe the same line, they have infinitely many solutions, as every point on the line satisfies both equations. This system is dependent and consistent because there is no contradiction between the equations.

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Key Concept: Coinciding Lines → Infinite Solutions

c) Infinitely many solutions

Solution Description: When two lines coincide, they represent infinitely many solutions. This occurs because the lines are essentially the same, meaning every point on one line is also on the other. In terms of a system of equations, this means the two equations describe the exact same relationship. Instead of intersecting at a single point or being parallel with no intersection, the lines overlap completely.

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Key Concepts:  Real-life Application

b) \$1.50
[Solution Description] Let the cost of a pencil be x dollars and the cost of an eraser be y dollars.
We have two equations: 3x + 5y = 12, 4x + 6y = 15
To solve using substitution or elimination, let’s eliminate y. Multiply the first equation by 4 and the second by 3 to make the cofficients of x equal:

12x + 20y = 48

12x + 18y = 45
Subtract the second equation from the first: 2y = 3 y = 1.5
Substitute y = 1.5 back into the first equation: 3x + 5(1.5) = 12

3x + 7.5 = 12

3x = 4.5

x = 1.5
Therefore, the cost of one pencil is \$1.

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Key Concepts: Solution Verification

(a) yes

Solution Description:

Since the point (1,2) satisfies both equations, it is a solution to the system of equations.

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Key Concept: Algebraic Dependence

b) Substitution Method

Solution Description: The algebraic method being used is substitution. In this approach, one equation is rearranged to express a variable in terms of another. This expression is then substituted into the second equation, eliminating one variable and allowing for a single-variable equation to be solved. Once the value of the variable is found, it is substituted back into the first equation to determine the remaining variable.

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Key Concept: Complex Elimination, Advanced Verification

a) $\frac{67} {51}$

Solution Description: To solve the system using elimination, first, multiply the equations by appropriate values so the coefficients of one variable match. Multiply the first equation by 3 so the coefficients of x become equal. Subtract one equation from the other to eliminate x and solve for y. Substitute the found value of y back into one of the original equations to determine x. Simplifying the equation gives the value of x. The elimination method helps remove one variable, making it easier to solve for the other. Finally, checking the solution ensures the values satisfy both original equations.

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Key Concept: Age Problem

a) Mary’s age: 45, Son’s age: 25

[Solution Description]

Let m be Mary’s current age and s her son’s current age.

From the problem statement: $m – 5 = 2(s – 5)$

Also given: $m = s + 20$

Substitute $m = s + 20$ into the first equation: $s + 20 – 5 = 2(s – 5)$

Simplify and solve for s:

$$s + 15 = 2s – 10$$

$$15 + 10 = 2s – s$$

$$s = 25$$

Substitute back to get m: $m = 25 + 20 = 45$

Hence, Mary is 45 years old and her son is 25 years old.

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