calendarBack
Quant

/

Algebra

/

Linear Equations
ALL MODULES

CAT 2025 Lesson : Linear Equations - Solving Equations

bookmarked

3.2 Solving Linear Equations

You might already be knowing how to solve linear equations. For many of us it flows subconsciously. Nevertheless, let us try to understand the process(es) to be followed.

3.2.1 Solving one variable

11) We keep the variable term(s) on one side and move the constant terms to the other.
22) We then move the coefficient of the variable to the other side.

For instance, where the equation is
3x+3=18−2x3x + 3 = 18 - 2x
⇒ 3x+2x=18−3 3x + 2x = 18 - 3
⇒ 5x=15 5x = 15

⇒ x=155=3 x = \dfrac{15}{5} = 3

3.2.2 Solving two variables

We can follow either of the following approaches. Most commonly used for general applicability is the Subtraction Method.

When we multiply or divide both sides of an equation by the same constant, the equation does not change. We use this in the Subtraction Method, which is outlined below

Subtraction Method

11) After identifying a variable to be eliminated, find the LCM of their coefficients in the 22 equations, ignoring the positive/negative sign.
22) We multiply or divide the entire equations by constants such that the LCM becomes the coefficients of the selected variable.
33) We now add or subtract the two equations (depending on the sign) to eliminate 11 variable.
44) We are now left with 11 equation containing only 11 variable. Upon solving this variable, we substitute this in either of the two equations to find the value of the other variable.

Substitution Method

11) In one of the equations, express one variable in terms of the other.
22) Substitute this in the other equation to get 11 equation with 11 variable.
33) Solve for this and substitute in either equation to find the other variable.

Note: In either of the above approaches, if both variables get eliminated when in Step 33 of either of the above approaches, then the equations could be dependent or inconsistent.

Example 4

Solve for the variables xx and yy if 3x+4y=103x + 4y = 10 and 2x+3y=72x + 3y = 7.

Solution

3x+4y=103x + 4y = 10 -----(11)
2x+3y=72x + 3y = 7 -----(22)

Subtraction Method

We can look to remove the xx-term from these equations. The coefficients of xx are 22 and 33, and their LCM is 66. We multiply each equation by a value such that the coefficient of the xx-term becomes 66.

Eq(11) ×2\times 2 ⇒ 6x+8y=206x + 8y = 20 -----(33)
Eq(22) ×3\times 3 ⇒ 6x+9y=216x + 9y = 21 -----(44)

Now if we subtract the equations, the x-terms are cancelled.
Eq(44) −- Eq(33) =(6x+9y)−(6x+8y)=21−20= (6x + 9y) - (6x + 8y) = 21 - 20
⇒ y=1 \bm{y = 1}

Substituting y=1y = 1 in Eq(11), we get
3x+4×1=103x + 4 \times 1 = 10
⇒ x=2 \bm{x = 2}

Substitution Method

We express one variable in terms of another and substitute in the other equation.
Eq(11)⇒ 3x+4y=10 3x + 4y = 10

x=10−4y3x = \dfrac{10 - 4y}{3}

Substituting this in Eq(22),
2×(10−4y3)+3y=72 \times \left(\dfrac{10 - 4y}{3}\right) + 3y = 7

⇒ 20−8y+9y3=7 \dfrac{20 - 8y + 9y}{3} = 7

⇒ 20+y=21 20 + y = 21
y=1\bm{y = 1}

Substituting y=1y = 1 in Eq (11), we get x=2\bm{x = 2}

Answer: (2,1)(2, 1)


3.2.3 Solving three Variables

We use the elimination approach here.
11) Follow the elimination approach as detailed above to eliminate 11 variable from 22 pairs of
equations, say equations A and B and equations B and C.
22) We now have 22 equations with the same 22 variables each.
33) We solve this as explained under the eliminations approach.

Example 5

Solve the variables if x+2y+3z=20x + 2y + 3z = 20, 2x+y+2z=152x + y + 2z = 15 and 3x+3y+2z=233x + 3y + 2z = 23

Solution

x+2y+3z=20x + 2y + 3z = 20 -----(11)
2x+y+2z=152x + y + 2z = 15 -----(22)
3x+3y+2z=233x + 3y + 2z = 23 -----(33)
The z coefficient is same in Eq(22) and Eq(33), so we shall look to remove the zz terms from 22 sets of equations.

Eq(33) - Eq(22) ⇒ x+2y=8 x + 2y = 8 -----(55)

Eq(22) ×3−\times 3 - Eq(11) ×2\times 2 ⇒ 4x−y=54x - y = 5 -----(66)

Eq(55) + 2×2 \times Eq(66) ⇒ 9x=18 9x = 18
⇒ x=2 \bm{x = 2}

Substituting x=2x = 2 in Eq(55), we get y=3\bm{y = 3}

Substituting x=2x = 2 and y=3y = 3 in Eq (33) we get
3×2+3×3+2×z=233 \times 2 + 3 \times 3 + 2 \times z = 23
⇒ z=4\bm{ z = 4}

∴(x,y,z)=(2,3,4)\therefore (x, y, z) = (2, 3, 4)

Answer: (2,3,4)(2, 3, 4)


Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock