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Linear Equations

Linear Equations

MODULES

Basics of Equations
Graph, Dependent & Inconsistent Equations
Solving Equations
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2 Equations & 3 Variables
Items Measured in Groups of 2 or more
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Linear Equations 1
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Linear Equations 2
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PRACTICE

Linear Equations : Level 1
Linear Equations : Level 2
Linear Equations : Level 3
ALL MODULES

CAT 2025 Lesson : Linear Equations - Solving Equations

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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

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

For instance, where the equation is
3x+3=18−2x3x + 3 = 18 - 2x3x+3=18−2x
⇒ 3x+2x=18−3 3x + 2x = 18 - 33x+2x=18−3
⇒ 5x=15 5x = 155x=15

⇒ x=155=3 x = \dfrac{15}{5} = 3x=515​=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

111) After identifying a variable to be eliminated, find the LCM of their coefficients in the 222 equations, ignoring the positive/negative sign.
222) We multiply or divide the entire equations by constants such that the LCM becomes the coefficients of the selected variable.
333) We now add or subtract the two equations (depending on the sign) to eliminate 111 variable.
444) We are now left with 111 equation containing only 111 variable. Upon solving this variable, we substitute this in either of the two equations to find the value of the other variable.

Substitution Method

111) In one of the equations, express one variable in terms of the other.
222) Substitute this in the other equation to get 111 equation with 111 variable.
333) 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 333 of either of the above approaches, then the equations could be dependent or inconsistent.

Example 4

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

Solution

3x+4y=103x + 4y = 103x+4y=10 -----(111)
2x+3y=72x + 3y = 72x+3y=7 -----(222)

Subtraction Method

We can look to remove the xxx-term from these equations. The coefficients of xxx are 222 and 333, and their LCM is 666. We multiply each equation by a value such that the coefficient of the xxx-term becomes 666.

Eq(111) ×2\times 2×2 ⇒ 6x+8y=206x + 8y = 206x+8y=20 -----(333)
Eq(222) ×3\times 3×3 ⇒ 6x+9y=216x + 9y = 216x+9y=21 -----(444)

Now if we subtract the equations, the x-terms are cancelled.
Eq(444) −-− Eq(333) =(6x+9y)−(6x+8y)=21−20= (6x + 9y) - (6x + 8y) = 21 - 20=(6x+9y)−(6x+8y)=21−20
⇒ y=1 \bm{y = 1}y=1

Substituting y=1y = 1y=1 in Eq(111), we get
3x+4×1=103x + 4 \times 1 = 103x+4×1=10
⇒ x=2 \bm{x = 2}x=2

Substitution Method

We express one variable in terms of another and substitute in the other equation.
Eq(111)⇒ 3x+4y=10 3x + 4y = 103x+4y=10

x=10−4y3x = \dfrac{10 - 4y}{3}x=310−4y​

Substituting this in Eq(222),
2×(10−4y3)+3y=72 \times \left(\dfrac{10 - 4y}{3}\right) + 3y = 72×(310−4y​)+3y=7

⇒ 20−8y+9y3=7 \dfrac{20 - 8y + 9y}{3} = 7320−8y+9y​=7

⇒ 20+y=21 20 + y = 2120+y=21
y=1\bm{y = 1}y=1

Substituting y=1y = 1y=1 in Eq (111), we get x=2\bm{x = 2}x=2

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


3.2.3 Solving three Variables

We use the elimination approach here.
111) Follow the elimination approach as detailed above to eliminate 111 variable from 222 pairs of
equations, say equations A and B and equations B and C.
222) We now have 222 equations with the same 222 variables each.
333) We solve this as explained under the eliminations approach.

Example 5

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

Solution

x+2y+3z=20x + 2y + 3z = 20x+2y+3z=20 -----(111)
2x+y+2z=152x + y + 2z = 152x+y+2z=15 -----(222)
3x+3y+2z=233x + 3y + 2z = 233x+3y+2z=23 -----(333)
The z coefficient is same in Eq(222) and Eq(333), so we shall look to remove the zzz terms from 222 sets of equations.

Eq(333) - Eq(222) ⇒ x+2y=8 x + 2y = 8x+2y=8 -----(555)

Eq(222) ×3−\times 3 -×3− Eq(111) ×2\times 2×2 ⇒ 4x−y=54x - y = 54x−y=5 -----(666)

Eq(555) + 2×2 \times2× Eq(666) ⇒ 9x=18 9x = 189x=18
⇒ x=2 \bm{x = 2}x=2

Substituting x=2x = 2x=2 in Eq(555), we get y=3\bm{y = 3}y=3

Substituting x=2x = 2x=2 and y=3y = 3y=3 in Eq (333) we get
3×2+3×3+2×z=233 \times 2 + 3 \times 3 + 2 \times z = 233×2+3×3+2×z=23
⇒ z=4\bm{ z = 4}z=4

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

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


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