Linear Equations Concepts

1. Simultaneous Multi-Variable Solving

• Formula: For

$$a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2$$

use elimination, substitution, or

$$x=\frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1},\qquad y=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}.$$

Use when: Two unknown quantities are connected by two linear conditions.

• Example: Solve (3x+2y=16) and (x-y=1).

Solution:

$$x-y=1\Rightarrow x=y+1$$

Substitute:

$$3(y+1)+2y=16$$ $$5y+3=16\Rightarrow y=\frac{13}{5}$$ $$x=y+1=\frac{18}{5}$$

Therefore,

$$\boxed{x=\frac{18}{5},\ y=\frac{13}{5}}$$

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2. Word-to-Equation Modeling

• Formula: Translate each condition into an equation. Common models:

$$\text{Sum}=x+y,\qquad \text{Difference}=x-y$$ $$\text{Total cost}=px+qy$$ $$\text{Total quantity}=x+y$$

Use when: A word problem gives two unknown quantities and two independent conditions.

• Example: The sum of two numbers is 30 and their difference is 4. Find the numbers.

Solution: Let the numbers be (x) and (y).

$$x+y=30$$ $$x-y=4$$

Adding:

$$2x=34\Rightarrow x=17$$ $$y=30-17=13$$ $$\boxed{17,\ 13}$$

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3. Cross-Multiplication for Two Linear Equations

• Formula: For

$$ax+by+c=0,\qquad dx+ey+f=0$$ $$x=\frac{bf-ce}{ae-bd},\qquad y=\frac{cd-af}{ae-bd}.$$

Use when: The coefficients are simple and direct substitution/elimination is less convenient.

• Example: Solve

$$2x+3y=12,\qquad x-y=1.$$

Solution:

Rewrite:

$$2x+3y-12=0$$ $$x-y-1=0$$

Here,

$$a=2,\ b=3,\ c=-12$$ $$d=1,\ e=-1,\ f=-1$$ $$x=\frac{bf-ce}{ae-bd} =\frac{3(-1)-(-12)(-1)}{2(-1)-3(1)}$$ $$x=\frac{-3-12}{-2-3}=3$$

Similarly,

$$y=\frac{cd-af}{ae-bd} =\frac{(-12)(1)-2(-1)}{-5} =2$$ $$\boxed{x=3,\ y=2}$$

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4. Ratio/Proportion-Based Linear Equations

• Formula: If

$$x:y=m:n,$$

write

$$x=mk,\qquad y=nk.$$

Use when: A problem gives a ratio along with a sum, difference, total cost, or another linear condition.

• Example: Two numbers are in the ratio (3:5). Their sum is 64. Find the numbers.

Solution: Let

$$x=3k,\qquad y=5k$$

Given:

$$3k+5k=64$$ $$8k=64\Rightarrow k=8$$

Therefore,

$$x=24,\qquad y=40$$ $$\boxed{24,\ 40}$$

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5. Geometry Application Models

• Formula: Common linear geometry relations:

$$P_{\text{rectangle}}=2(l+b)$$ $$P_{\text{square}}=4a$$ $$l=b+d \quad\text{(if length exceeds breadth by (d))}$$

Use when: Geometric information creates simultaneous linear equations.

• Example: A rectangle has perimeter (40) m. Its length is (4) m more than its breadth. Find its dimensions.

Solution: Let breadth (=b).

$$l=b+4$$

Perimeter:

$$2(l+b)=40$$ $$l+b=20$$

Substitute:

$$b+4+b=20$$ $$2b=16\Rightarrow b=8$$ $$l=12$$ $$\boxed{l=12\text{ m},\ b=8\text{ m}}$$

        12 m
   ┌────────────┐
8m │            │
   └────────────┘
        12 m

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6. Mixture & Cost Equations

• Formula:

$$\text{Total quantity}=x+y$$ $$\text{Total cost}=c_1x+c_2y$$

Hence,

$$\text{Average price}=\frac{c_1x+c_2y}{x+y}.$$

Use when: Two quantities with different prices/concentrations are mixed.

• Example: Rice costing ₹40/kg and ₹60/kg is mixed to obtain (20) kg at ₹48/kg. Find the quantity of each.

Solution: Let cheaper rice (=x) kg.

Expensive rice:

$$20-x$$

Total cost:

$$40x+60(20-x)=48(20)$$ $$40x+1200-60x=960$$ $$-20x=-240$$ $$x=12$$

Therefore expensive rice:

$$20-12=8$$ $$\boxed{12\text{ kg at ₹40/kg},\ 8\text{ kg at ₹60/kg}}$$

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Advanced Variants

7. Parameter Conditions for Unique, No, or Infinite Solutions

• Formula: For

$$a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2,$$

unique solution exists when

$$a_1b_2-a_2b_1\ne0.$$

If

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2},$$

there is no solution. If all three ratios are equal, there are infinitely many solutions.

• Example: Find (k) so that

$$2x+3y=5,\qquad 4x+ky=10$$

has infinitely many solutions.

Solution: For infinitely many solutions:

$$\frac{2}{4}=\frac{3}{k}=\frac{5}{10}$$ $$\frac{1}{2}=\frac{3}{k}$$ $$k=6$$ $$\boxed{k=6}$$

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8. Three-Variable Linear Systems

• Formula: For three equations in (x,y,z), eliminate one variable from two pairs of equations, then solve the resulting two-variable system. Use when: Aptitude problems involve three unknown quantities such as prices, ages, or quantities.

• Example: Solve:

$$x+y+z=9$$ $$x+y=5$$ $$y+z=6.$$

Solution: From the first two:

$$z=9-5=4$$

From (y+z=6):

$$y+4=6\Rightarrow y=2$$

Then:

$$x+y=5\Rightarrow x=3$$ $$\boxed{x=3,\ y=2,\ z=4}$$