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Pair of Linear Equations in Two Variables - Model and solve situational word problems using simultaneous 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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To model situational problems: First, identify the two unknown quantities and assign them variables (usually xx and yy).

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Translate the given verbal conditions into two distinct algebraic equations.

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Common types include: Age problems (using x−nx - n for past and x+nx + n for future), Digit problems (Number represented as 10x+y10x + y), and Speed-Distance problems.

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For boat problems: If the speed of the boat in still water is xx km/h and the speed of the stream is yy km/h, then Speed Downstream = (x+y)(x + y) km/h and Speed Upstream = (x−y)(x - y) km/h.

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Equations are solved using methods like Elimination or Substitution to find the values of the variables.

📐Formulae

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

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

Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

Number=10(tens digit)+(units digit)\text{Number} = 10(\text{tens digit}) + (\text{units digit})

vdownstream=x+yv_{downstream} = x + y

vupstream=x−yv_{upstream} = x - y

💡Examples

Problem 1:

The sum of the digits of a two-digit number is 99. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the digit at the tens place be xx and the digit at the units place be yy.

The number is 10x+y10x + y. When digits are reversed, the new number is 10y+x10y + x.

According to the first condition: x+y=9…(1)x + y = 9 \quad \dots(1)

According to the second condition: 9(10x+y)=2(10y+x)9(10x + y) = 2(10y + x) 90x+9y=20y+2x90x + 9y = 20y + 2x 88x−11y=088x - 11y = 0 Dividing by 1111: 8x−y=0…(2)8x - y = 0 \quad \dots(2)

Adding equations (1)(1) and (2)(2): x+y=9+8x−y=09x=9\begin{array}{r} x + y = 9 \\ + \quad 8x - y = 0 \\ \hline 9x = 9 \end{array} x=1x = 1

Substituting x=1x = 1 in (1)(1): 1+y=9⇒y=81 + y = 9 \Rightarrow y = 8

The number is 10(1)+8=1810(1) + 8 = 18.

Explanation:

We define variables for the digits, then construct equations based on the sum of digits and the relationship between the original and reversed numbers. We solve the system using the elimination method.

Problem 2:

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27₹ 27 for a book kept for seven days, while Susy paid ₹21₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first 33 days be ₹x₹ x and the additional charge per day be ₹y₹ y.

For Saritha (7 days total = 3 days fixed + 4 days extra): x+4y=27…(1)x + 4y = 27 \quad \dots(1)

For Susy (5 days total = 3 days fixed + 2 days extra): x+2y=21…(2)x + 2y = 21 \quad \dots(2)

Subtracting (2)(2) from (1)(1): x+4y=27−(x+2y=21)2y=6\begin{array}{r} x + 4y = 27 \\ - (x + 2y = 21) \\ \hline 2y = 6 \end{array} y=3y = 3

Substituting y=3y = 3 in (2)(2): x+2(3)=21x + 2(3) = 21 x+6=21x + 6 = 21 x=15x = 15

The fixed charge is ₹15₹ 15 and the charge per extra day is ₹3₹ 3.

Explanation:

Identify the two types of costs: fixed and variable. Translate the scenarios into linear equations by subtracting the initial 3-day fixed period from the total days to find the variable portion. Solve via elimination.