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Linear Equations in Two Variables - Form and solve pairs of linear equations in two variables from contexts

Grade 9CBSE

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

🔑Concepts

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A linear equation in two variables is an equation that can be put in the form ax+by+c=0ax + by + c = 0, where a,b,ca, b, c are real numbers and aa and bb are not both zero.

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A 'pair' of linear equations (system of equations) represents two relationships between the same two variables xx and yy.

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To form equations from context, identify the two unknown quantities and assign them variables (usually xx and yy). Then, translate the given conditions into mathematical expressions.

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Common word problem triggers: 'Sum' means ++, 'Difference' means −-, 'Times' or 'Product' means ×\times, and 'Is' or 'Results in' means ==.

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A solution to a pair of linear equations is a pair of values (x,y)(x, y) that satisfies both equations simultaneously.

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Graphically, the solution to a pair of linear equations is the point (x,y)(x, y) where the two straight lines intersect.

📐Formulae

ax+by+c=0ax + by + c = 0

{a1x+b1y+c1=0a2x+b2y+c2=0\begin{cases} a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0 \end{cases}

Total Cost=(Price per unit×Quantity A)+(Price per unit×Quantity B)\text{Total Cost} = (\text{Price per unit} \times \text{Quantity A}) + (\text{Price per unit} \times \text{Quantity B})

💡Examples

Problem 1:

The sum of two numbers is 4545 and their difference is 77. Form a pair of linear equations and find the numbers.

Solution:

Let the first number be xx and the second number be yy. Based on the first condition: x+y=45x + y = 45 (Eq. 1) Based on the second condition: x−y=7x - y = 7 (Eq. 2)

To solve, add Eq. 1 and Eq. 2: x+y=45+(x−y=7)2x=52\begin{array}{r} x + y = 45 \\ + (x - y = 7) \\ \hline 2x = 52 \end{array} x=522=26x = \frac{52}{2} = 26

Substitute x=26x = 26 into Eq. 1: 26+y=4526 + y = 45 y=45−26=19y = 45 - 26 = 19

The numbers are 2626 and 1919.

Explanation:

We translated the sentences into two algebraic equations. By adding them, we eliminated yy to solve for xx, then substituted xx back to find yy.

Problem 2:

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Form the pair of linear equations to represent this situation.

Solution:

Let Nuri's present age be xx years and Sonu's present age be yy years.

Five years ago: Nuri's age =x−5= x - 5 Sonu's age =y−5= y - 5 Equation: x−5=3(y−5)⇒x−5=3y−15⇒x−3y+10=0x - 5 = 3(y - 5) \Rightarrow x - 5 = 3y - 15 \Rightarrow x - 3y + 10 = 0

Ten years later: Nuri's age =x+10= x + 10 Sonu's age =y+10= y + 10 Equation: x+10=2(y+10)⇒x+10=2y+20⇒x−2y−10=0x + 10 = 2(y + 10) \Rightarrow x + 10 = 2y + 20 \Rightarrow x - 2y - 10 = 0

The pair of equations is: x−3y=−10x - 3y = -10 x−2y=10x - 2y = 10

Explanation:

Age problems require adjusting the current age (xx and yy) by subtracting for the past or adding for the future before applying the 'thrice' or 'twice' multipliers.

Problem 3:

The cost of 22 pencils and 33 erasers is ₹9\text{₹} 9 and the cost of 44 pencils and 66 erasers is ₹18\text{₹} 18. Represent this situation algebraically.

Solution:

Let the cost of one pencil be ₹x\text{₹} x and the cost of one eraser be ₹y\text{₹} y.

From the first condition: 2x+3y=92x + 3y = 9

From the second condition: 4x+6y=184x + 6y = 18

Note that the second equation is just the first equation multiplied by 22. This indicates that the lines are coincident and there are infinitely many solutions.

Explanation:

We assign variables to the unit costs and create total cost equations. Because the coefficients are proportional (24=36=918\frac{2}{4} = \frac{3}{6} = \frac{9}{18}), these equations represent the same line.