Linear Equations
An equation is a mathematical statement which implies that two quantities separated by the symbol ‘=’ are equal.
For example,
3x + 6 = 0,
x2 + 5x = 
6
The terms on the left hand side and the right hand side of the ‘=’ sign are called the LHS and RHS respectively.
Equations contain variables, coefficients and constants. A variable is the unknown quantity in the equation. Variables are generally denoted by letters like x, y, z, a, b, c etc. An equation can contain one or more variables. Coefficient is the number which multiplies the variable and a constant is an independent numerical value in the equation. For instance, in the equation 3x + 6 = 0, x is the variable, 3 is the coefficient of the variable and 6 is the constant.
Depending on the highest power of the variable(s), equations are classified as linear, quadratic, cubic etc. The highest power of the variable(s) in a linear equation is 1 and that in a quadratic equation is 2, for a cubic equation is 3 and so on. In this chapter, we will study linear equations in detail.
To solve an equation is to find a numerical value that the variable can take so that the equation holds true. Such a value of the variable is called the solution.
For instance, consider x + 4 = 8. This equation can hold true only if x = 4. Thus, x = 4 is a solution of the equation.
Similarly, if 3x + 6 = 0, then
x = 2
is a solution of the equation.
REMEMBER
- The solution of a given equation does not change if we add or subtract the same number from both sides of the equation, nor does it change if we multiply both the sides with the same number or divide both the sides with the same non-zero number.
- To solve a linear equation in one variable the following steps should be followed:
Step 1: Simplify the LHS and the RHS by removing brackets.
Step 2: Shift all the terms containing the variable to the LHS and the constant terms to the RHS with appropriate change of sign and then simplify.
While shifting terms to the opposite side, a positive sign changes to negative (and vice versa) and a multiplication changes to a division.
|
Find the value of x, if 7x + 5(2
Solution:
Step 1:
Step 2:
|
|
Solution: Step 1:
Step 2:
|
Forming a linear equation from the given information is as important a step as solving it. Let us look at a few examples.
REMEMBER
- During the exam, some multiple choice questions can be solved by substituting the variable with the values given in the options or substituting values in place of variables like in the following example.
|
[CAT 2002]
Solution:
Case 1: Let p = 2/3, q = 3/2 and r = 1 Substituting the values of p, q, r, we get,
Case 2: Let p = 1, q = 1 and r = 1 Substituting the values of p, q, r, we get,
Hence, option 3. |
pqr = 1
|
The sum of three consecutive numbers is 84. Find the numbers.
Solution: Let x be the first number. Then the other two consecutive numbers are x + 1 and x + 2. Hence we have, x + (x + 1) + (x + 2) = 84
The numbers are 27, 28 and 29. |
|
After purchasing four mangoes, a man commented: "Well, if the price of a mango was Rs. 4 less, I would have got two more mangoes for the same amount of money I have now paid". What is the price of one mango?
[JMET 2009]
(1) Rs. 8(2) Rs. 10 (3) Rs. 15(4) Rs. 12
Solution: Let the price of mango be x. If the price of mango was Rs. 4 less, I would have got two more mangoes for the same amount of money
Hence, option 4. |
|
Tickets for a concert were sold at Rs. 5, Rs. 3 and Re. 1 each. Thirty more tickets were sold at Rs. 5 than at Rs. 3, and twice as many at Re. 1 as at Rs. 3. If total receipts from the sale of tickets were Rs. 950, then how many tickets of each kind were sold?
Solution: Let the number of Rs. 3 tickets = x Number of Rs. 5 tickets = 30 + x Number of Re. 1 tickets = 2x Total amount = Rs. 950
|
|
A string of length 40 metres is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 metres smaller than the first part. Find the length of the largest part. [CAT 2002] (1) 27(2) 4 (3) 5(4) 9
Solution: Let l, m and s be the longest, medium and the shortest lengths of the strings. l = 3m and s = l – 23 l + s + m = 40
l = 27 Hence, option 1. |
|
A man received a cheque. The amount in Rs. has been transposed for paise and vice versa. After spending Rs. 5 and 42 paise, he discovered he now had exactly 6 times the value of the correct cheque amount. What amount he should have received? [CAT 2002] (1) Rs. 5.30(2) Rs. 6.44 (3) Rs. 60.44(4) Rs.6.44
Solution: Let the man has received a cheque of x rupees and y paise.
The amount actually received by him = 100y + x After spending Rs. 5 and 42 paise, the remaining amount = (100y + x – 542 …(ii) But, (100y + x – 542) = 6 Substituting the values from the given options, x = 6 and y = 44 Hence, option 2. |
Now let us try to understand a linear equation graphically. A linear equation of the form ax + by = c, where x and y are variables, a and b are coefficients and c is a constant, represents a line on the X–Y plane. Hence, any equation of this form can be plotted on the X–Y plane. A linear equation when plotted on a graph will give a straight line. Graphical representation of linear equations would enable us to understand linear equations in a better way.
A linear equation with only one variable gives a line which is parallel to the x-axis or the y-axis.
For example, x – 2 = 0 can be represented as:
The solution of the linear equation is the point where the line intersects the axis. In this case the solution is x = 2.
An equation of the form ax + by = c, where the highest power of the variables x and y is unity is called a linear equation in two variables. Here a ≠ 0, b ≠ 0 and a, b and c are real numbers. The values of x and y for which the equation holds are called the solution of the equation.
A line representing a linear equation with two variables cannot be parallel to the co-ordinate axes.
For example, let us represent the equation
3x + 4y = 12 on the X–Y plane.
We have,
3x + 4y = 12 …(i)
Put x = 0, then equation (i) becomes
0 + 4y = 12
4y = 12
y = 3
Thus (0, 3) are the co-ordinates of a point on the line.
Put y = 0, then equation (i) becomes
3x + 0 = 12
3x = 12
x = 4
Hence (4, 0) are the co-ordinates of another point on the same line.
Hence we can say that (0, 3) and (4, 0) are the Y - intercept and X - intercept of the line respectively. We join these two points on the X–Y plane as shown and the line thus formed is the graphical representation of the linear equation 3x + 4y = 12.
All the points on the line shown satisfy this equation. A line extends in both directions infinitely. Hence such an equation has infinite solutions.
REMEMBER
- A linear equation with two or more variables does not have a unique solution. It has an infinite number of solutions.
EQUATIONS IN TWO VARIABLES WITH POSITIVE INTEGER SOLUTIONS
Any one linear equation in two variables can have an infinite number of solutions, but if the solutions are known to be positive integers, the number of solutions can be finite and can be determined as shown in the following example.
|
Find how many positive integers x and y satisfy the equation 5x + 3y = 59.
Solution: 5x + 3y = 59
As x and y are positive integers,
y Thus the smallest possible value of y is 3. Corresponding value of x is 10. The other possible values of y are 8, 13 and 18, and the corresponding values of x are 7, 4 and 1. x is negative for higher possible values of y. Thus there are 4 possible solutions. |
1 and (4 – 3y)/5 is an integer.
|
A test has 50 questions. A student scores 1 mark for
a correct answer, [CAT 2003 Leaked Test] (1) 6(2) 12 (3) 3(4) 9 Solution: Let R, W and N be the number of questions with right answers, questions with wrong answers and not attempted questions respectively. From the conditions given in the question, we have, R + W + N = 50… (i) R Solving equations (i) and (ii), we get, 7R
Hence, option 3. |
REMEMBER
- For an equation ax – by = c (a, b and c are positive integers), having positive integer solutions, the values of x form an arithmetic progression with common difference b and the values of y form an arithmetic progression with common difference a.
Linear equations in two variables, which are both satisfied by the same unique solution, are called simultaneous equations. Graphically, these equations are two non parallel lines which intersect at a particular point (x, y), which is the solution of both the equations.
For example, 3x + y = 5 and 5x – y = 3 intersect at x = 1 and y = 2.
Thus x = 1 and y = 2 lies on both the lines and we say that the solution of the equations 3x + y = 5 and 5x – y = 3 is (1, 2). In other words (1, 2) is a point that lies on both the lines. Hence the point should satisfy both the equations.
Substituting x = 1 and y = 2 in 3x + y = 5 we have,
LHS = 3x + y
= 3(1) + 2
= 5
= RHS
Similarly, substituting x = 1 and y = 2 in the equation 5x – y = 3 we have,
LHS = 5x – y
= 5(1) – 2
= 3
= RHS
In the above example, we can see that 5 – 3x > 5x – 3, for x < 1 and 5 – 3x < 5x – 3, for x > 1. Also, we note that the minimum possible value that the greater value of (5 – 3x) and (5x – 3) can take is at x = 1 and this value is 2.
Two or more linear equations in the same number of variables having a common solution are called a system of simultaneous equations. The common solutions to these equations can be found out by graphical method as explained earlier or by the algebraic method discussed below.
- SOLVING SIMULTANEOUS EQUATIONS
It is impractical to plot equations on a graph paper to find common solutions. Hence, we use the algebraic method to obtain the same common solutions of simultaneous equations.
Two general methods of solving simultaneous equations are described below:
SUBSTITUTION METHOD
Consider the equations 3y
x = 1 and 7y
2x = 4 to be solved simultaneously. Thus we have,
3y – x = 1…(i)
7y – 2x = 4…(ii)
From equation(i),
3y = 1 + x
Substituting this value of y in equation (ii),
7 + 7x – 6x = 12
x = 5
On substituting x = 5 in equation (iii), we have
y = 2
x = 5 and y = 2 satisfy the two equations and (5, 2) is a solution
of the two equations.
ELIMINATION METHOD
Consider the equations 2x + 3y = 4 and 3x + 4y = 5 to be solved simultaneously. Thus we have,
2x + 3y = 4 …(i)
3x + 4y = 5 …(ii)
Multiply each equation by the coefficient of x (or y) in the other equation.
Multiplying equation (i) by 3 and equation (ii) by 2
6x + 9y = 12…(iii)
6x + 8y = 10…(iv)
Subtracting equation (iv) from equation (iii),
y = 2
Substituting y = 2 in equation (i)
2x + 3 
2 = 4
x = –2/2 = –1
x = –1 and y = 2 satisfy both the equations.
(–1, 2) is a solution of the two equations.
|
In a particular jungle which only had deer and human visitors, there were 70 heads and 188 legs. How many deer and visitors were there?
Solution: Let the number of deer be x and the number of visitors be y. x + y = 70...(i) Since deer have 4 legs and human visitors have 2 legs we have, 4x + 2y = 188...(ii) Multiplying equation (i) by 2 and subtracting the same from equation (ii), we get x = 24 and y = 46 |
|
The sum of the digits of a two digit number is 7. If the digits are reversed, the number so obtained when increased by 3 equals 4 times the original number. Find the original number.
Solution: Let the digit in the ten’s place = x Let the digit in the unit’s place = y So the number = 10x + y If the digits are reversed, the new number = 10y + x According to given conditions, x + y = 7...(i) (10y + x) + 3 = 4 (10x + y)
Dividing the above equation by –3
Multiplying equation (i) by 2 and adding equation (ii) 15x = 15
Substituting the value of x in equation (i), we get y = 6
|
- INDETERMINATE SIMULTANEOUS EQUATIONS
Two given equations that can be algebraically derived from each other are called dependent equations. Such equations have infinitely many solutions and are called Indeterminate Simultaneous Equations. For example,
4x + 5y = 18…(i)
8x + 10y = 36…(ii)
The above two equations are dependent equations as equation (ii) can be obtained by multiplying equation (i) with 2.
In general, simultaneous equations ax + by + c = 0 and lx + my + n = 0 are called indeterminate if and only if
|
For what value of k will the following system of equations be indeterminate? 2x – 3y = 5 6x + ky = 15
Solution: Here, the ratio of the coefficients of x, which is 1/3, is the same as the ratio of the constant terms in the two equations. Hence, for the equations to be indeterminate, the ratio of coefficients of y also would be 1/3.
Hence, the value of k for which the equations are indeterminate is – 9. |
REMEMBER
- Indeterminate simultaneous equations have infinite number of solutions. Graphically, they are two lines superimposed on each other.
- INCONSISTENT SIMULTANEOUS EQUATIONS
Inconsistent equations are those, for which there is no common solution, i.e. there is not a single pair of numbers that satisfies both the equations simultaneously, although each of the equations has infinite solutions. For example,
13x + 32y = 19…(i)
26x + 64y = 34…(ii)
Multiplying equation (i) by 2, we get 26x + 64y = 38, which contradicts equation (ii). Thus equation (i) and equation (ii) are not indeterminate but are inconsistent.
In general two equations of the form ax + by + c = 0 and lx + my + n = 0 are inconsistent if and only if
REMEMBER
- Graphically, inconsistent simultaneous equations are two parallel lines in a plane, which can never intersect each other.
|
Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0? x + 2y 2x + 6y x [CAT 2003 Leaked Test] (1) 5p (3) 5p + 2q
Solution: On substituting the values of p, q and
r in the options we see that the values of p, q and
r satisfy only the equation 5p Hence, option 1. |
A set of n unique equations (those which cannot be algebraically derived from one another) in n variables also can be solved simultaneously for a common solution.
For example,
We have the following set of equations,
x + y + 2z = 9 ...(i)
2x + 4y + 6z = 28...(ii)
3x + 4y + 5z = 26...(iii)
The above set of equations is a system of three equations with three variables.
Both substitution and elimination methods can be used to solve these equations.
If we multiply (i) by 2 and subtract it from (ii), we get
y + z = 5...(iv)
If we multiply (i) by 3 and subtract it from (iii), we get
z – y = 1...(v)
Equation (iv) and equation (v) can be added to get z = 3.
Using this value of z we can easily find y and x.
REMEMBER
- To solve a system of simultaneous equations, the number of independent equations must be at least equal to the number of variables.
|
Solve the system of equations for x, y and z. x – y + z = 6 2x + 2y + 3z = 11 2x – 3y – 2z = 1
Solution: x – y + z = 6...(i) 2x + 2y + 3z = 11...(ii) 2x – 3y – 2z = 1...(iii) Multiplying equation (i) by 2 and subtracting from equation (ii) we get, 4y + z = –1...(iv) Subtracting (iii) from (ii), 5y + 5z = 10 or y + z = 2...(v) Now equation (iv) and (v) are simultaneous equations in y and z. Solving, y = –1 and z = 3 Substituting the value of y and z in (i), we get x = 2
Thus x = 2, y = –1 and z = 3. |
|
Find the values of x, y and z. z + 4y + 3x = 33 4x + y – z = 6 2y + 8x – 2z = 12
Solution: z + 4y + 3x = 33...(i) 4x + y – z = 6 ...(ii) 2y + 8x – 2z = 12...(iii) Dividing equation (iii) by 2, y + 4x – z = 6, which is the same as equation (ii). Thus we have only two equations but three variables. Hence, the values of x, y and z cannot be found. |
|
Ina bought 2 candies, 3 cans of soft drink and 9 pieces of cake for Rs. 109. Sona bought 4 candies, 5 pieces of cake and a can of soft drink for Rs. 67. How much did Koena pay for 3 candies, 2 cans of soft drink and 7 pieces of cake, if all the three bought candies, cakes and soft drinks of the same kind?
Solution: Let a, b and c be the price of a candy, a can of soft drink and a piece of cake respectively. 2a + 3b + 9c = 109...(i) 4a + b + 5c = 67...(ii) Adding (i) and (ii), 6a + 4b + 14c = 176...(iii) Dividing equation (iii) by 2, 3a + 2b + 7c = 88 Thus, Koena paid Rs. 88 for 3 candies, 2 cans of soft drink and 7 pieces of cake. |
|
A change making machine contains 1 rupee, 2 rupee and 5 rupee coins. The total number of coins is 300. The amount is Rs. 960. If the number of 1 rupee coins and the number of 2 rupee coins are interchanged, the value comes down by Rs. 40. The total number of 5 rupee coins is [CAT 2001]
(1) 100(2) 140 (3) 60(4) 150 Solution: Let the number of Re. 1, Re. 2 and Re. 5 coins be x, y and z respectively. x + y + z = 300...(i) x + 2y + 5z = 960...(ii) 2x + y + 5z = 920...(iii) Adding equation (ii) and equation (iii), we get, 3x + 3y + 10z = 1880...(iv) Multiply equation (i) by 3, we get, 3x + 3y + 3z = 900...(v) Subtracting equation (v) from equation (iv), we get, 7z = 980
Hence, option 2. |
|
In a cricket match, Team A scored 232 runs without losing a wicket. The score consisted of byes, wides and runs scored by two opening batsmen: Ram and Shyam. The runs scored by the two batsmen are 26 times wides. There are 8 more byes than wides. If the ratio of the runs scored by Ram and Shyam is 6 : 7, then the runs scored by Ram is [XAT 2008] (1) 88(2) 96 (3) 102(4) 112 (5) None of these Solution: Let the number of runs scored by byes, wides and runs be x, y and z respectively.
Substituting equations (iii) and (ii) in equation (i), we get, y = 8
Let the runs scored by Ram be 6r and by Shyam be 7r.
Hence, option 2. |
|
An artist has completed one fourth of a rectangular oil painting. When he will paint another 100 square centimeters of the painting, he would complete three quarters of the painting. If the height of the oil painting is 10 centimeters, determine the length (in centimeters) of the oil painting. [JMET 2009] (1)15(2) 20 (3) 10(4) 25 Solution: Let the total area of the painting be x cm2.
As the canvas is rectangular in shape, Area of the canvas = Length
But, Height of the painting = 10 cm
Hence, option 2. |
|
A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x? [CAT 2008]
(1) 2 (3) 9 (5) 13 Solution: The store has x kg of rice, initially. (Henceforth, x is assumed to be measured in kg.) The first customer buys half the total rice in the store, and another half kg.
The second customer buys half of this, and another half kg. Amount purchased by second customer
The third customer also buys half the remaining rice, and another half kg. Amount purchased by third customer
Since we know that after this purchase, there was no more rice left in the store, we conclude that:
Hence, option 2. |
x 
|
In second year, students at a business school can opt for Systems, Operations or HR electives only. The number of girls opting for Operations and the number of boys opting for Systems elective is 37. Twenty two students opt for operations elective. Twenty girls opt for Systems and Operations electives. The number of students opting for Systems elective and the number of boys opting for Operations electives is 37. Twenty-five students opt for HR electives. [XAT 2008]
Question1: The number of students in the second year is? (1) 73(2) 74 (3) 75(4) 76 (5) 77 Solution:
From the given data we have, 20 – b + a = 37… (i) c + d = 37… (ii) a + b = d… (iii) c – b = 2… (iv) On solving these equations we get, a = 23, b = 6, c = 8 and d = 29
= 29 + 25 + 22 =76 Hence, option 4.
Question 2: If 20% of the girls opt for HR electives, then the total number of boys in the second year is (1) 54(2) 53 (3) 52(4) 51 (5) 50 Solution:
Now, y = 20% of total number of girls.
As total number of students = 76 and total number of girls = 25
Hence, option 4. |
