CBSE Class 10 Mathematics Simple Equations Notes Set 05

Simple (Linear) Equations

An equation of the first degree in one unknown is a simple or linear equation. To solve it, remove fractions by multiplying both sides by the L.C.M. of the denominators, remove brackets, collect the terms in \(x\) on one side and the numbers on the other, and divide by the coefficient of \(x\). The value found is the root, and it should always be verified in the original equation.

Example. Solve \(9x+5=6x+20\). Then \(3x=15\) and \(x=5\). Example. Solve \(\dfrac{3x-4}{4}-\dfrac{4x-1}{3}=\dfrac{x+2}{2}-6\). Multiplying by 12: \(9x-12-16x+4=6x+12-72\), so \(-7x-8=6x-60\), \(13x=52\) and \(x=4\).

Forming Equations from Word Problems

Let the unknown quantity be \(x\), write each statement of the problem in terms of \(x\), and equate the two expressions for the same quantity.

1. Ages. A father is 35 years old and his son 7. In how many years will the father be three times as old? \(35+x=3(7+x)\) gives \(x=7\). In 7 years they are 42 and 14.
2. Digits. A two-digit number has one digit double the other, and adding 18 reverses the digits. Let the smaller digit be \(x\), so the number is \(10x+2x=12x\) and the reversed number is \(21x\). \(12x+18=21x\) gives \(x=2\), the number is 24 and \(24+18=42\).
3. Wages and fines. A man is engaged for 70 days at Rs. 24 a day for working and pays Rs. 6 for each day he is absent. He receives Rs. 1230. If he worked \(x\) days: \(24x-6(70-x)=1230\), \(30x=1650\), \(x=55\).
4. Area. The length of a hall is twice its breadth. Adding 2 m to both increases the area by 52 m\(^2\). \((x+2)(2x+2)=2x^2+52\) gives \(6x+4=52\), \(x=8\), so the hall is 8 m by 16 m with area 128 m\(^2\).
5. Clock. The minute hand gains 55 minute spaces on the hour hand in an hour, that is 11/12 space per minute. The hands are together again when \(x-5=\dfrac{x}{12}\), so \(x=\dfrac{60}{11}=5\dfrac5{11}\) minutes past 1.

Simultaneous Equations in Two Variables

Two linear equations in \(x\) and \(y\) can be solved by elimination (make the coefficients of one variable equal and subtract or add), by substitution, or by cross-multiplication. Example: a car goes uphill at 30 km/h and downhill at 50 km/h and covers 650 km in 15 hours. If it goes uphill for \(x\) hours and downhill for \(y\) hours, \(30x+50y=650\) and \(x+y=15\). Substituting \(y=15-x\): \(30x+750-50x=650\), so \(x=5\) and \(y=10\).

Equations such as \(\dfrac{12}{x}-\dfrac{16}{y}=0\) become linear after putting \(p=1/x\) and \(q=1/y\).

Conditions for Solvability

For \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) the cross-multiplication result is \(\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}\).

Example: \(3x+5y=11\) and \(6x+10y=22\) have ratios \(\dfrac36=\dfrac5{10}=\dfrac{11}{22}\), so they have infinitely many solutions.