Linear Equations Problems and Solutions

Pair of Linear Equations in Two Variables

1-Mark Questions

For what value of k, the following system of equations has a unique solution: x - ky = 2 and 3x + 2y = -5 (M'15)

Step 1: Condition for unique solution
For a system of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have a unique solution:
a₁/a₂ ≠ b₁/b₂

Step 2: Apply the condition
For equations: x - ky = 2 and 3x + 2y = -5
a₁ = 1, b₁ = -k, a₂ = 3, b₂ = 2
So, 1/3 ≠ -k/2

Step 3: Solve for k
1/3 ≠ -k/2
Cross-multiplying: 2 ≠ -3k
k ≠ -2/3

∴ The system has a unique solution for all values of k except k = -2/3

For what values of m, the pair of equations 3x + my = 10 and 9x + 12y = 30 have a unique solution. (M'16)

Step 1: Condition for unique solution
For a unique solution: a₁/a₂ ≠ b₁/b₂

Step 2: Apply the condition
For equations: 3x + my = 10 and 9x + 12y = 30
a₁ = 3, b₁ = m, a₂ = 9, b₂ = 12
So, 3/9 ≠ m/12

Step 3: Solve for m
1/3 ≠ m/12
Cross-multiplying: 12 ≠ 3m
m ≠ 4

∴ The system has a unique solution for all values of m except m = 4

In a rectangle ABCD, AB = x + y, BC = x - y, CD = 9 and AD = 3. Find the values of x and y. (J'16)

Step 1: Properties of a rectangle
In a rectangle, opposite sides are equal.
So, AB = CD and BC = AD

Step 2: Set up equations
AB = CD ⇒ x + y = 9 ...(1)
BC = AD ⇒ x - y = 3 ...(2)

Step 3: Solve the equations
Adding (1) and (2):
(x + y) + (x - y) = 9 + 3
2x = 12 ⇒ x = 6

Step 4: Find y
Substitute x = 6 in equation (1):
6 + y = 9 ⇒ y = 3

∴ x = 6, y = 3

Show that the pair of Linear Equations 7x + y = 10 and x + 7y = 10 are consistent. (M'17)

Step 1: Check for consistency
For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
If a₁/a₂ ≠ b₁/b₂, the equations are consistent with a unique solution.

Step 2: Apply the condition
For equations: 7x + y = 10 and x + 7y = 10
a₁ = 7, b₁ = 1, a₂ = 1, b₂ = 7
a₁/a₂ = 7/1 = 7
b₁/b₂ = 1/7 = 1/7

Step 3: Compare the ratios
Since 7 ≠ 1/7, the equations are consistent with a unique solution.

∴ The pair of equations is consistent as a₁/a₂ ≠ b₁/b₂

Write the Condition for the pair of linear equations in two variables to be parallel lines. (J'17)

Condition for parallel lines:
For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
The lines are parallel if a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Explanation:
When a₁/a₂ = b₁/b₂, the lines have the same slope but different intercepts, making them parallel.

∴ The condition for parallel lines is: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

If x = a and y = b is solution for the pair of equations x - y = 2 and x + y = 4, then find the values of a and b. (M'18)

Step 1: Set up the equations
Since (a, b) is a solution to both equations:
a - b = 2 ...(1)
a + b = 4 ...(2)

Step 2: Solve the equations
Adding (1) and (2):
(a - b) + (a + b) = 2 + 4
2a = 6 ⇒ a = 3

Step 3: Find b
Substitute a = 3 in equation (2):
3 + b = 4 ⇒ b = 1

∴ a = 3, b = 1

Whether the following pair of Linear Equations are parallel? Justify. 6x - 4y + 10 = 0, 3x - 2y + 6 = 0. (J'18)

Step 1: Condition for parallel lines
For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 6x - 4y + 10 = 0: a₁ = 6, b₁ = -4, c₁ = 10
For 3x - 2y + 6 = 0: a₂ = 3, b₂ = -2, c₂ = 6

Step 3: Compare ratios
a₁/a₂ = 6/3 = 2
b₁/b₂ = -4/-2 = 2
c₁/c₂ = 10/6 = 5/3

Step 4: Check the condition
Since a₁/a₂ = b₁/b₂ = 2, but 2 ≠ 5/3

∴ Yes, the lines are parallel as a₁/a₂ = b₁/b₂ ≠ c₁/c₂

For what value of 't' the following pair of linear equations has no solution? 2x - ty = 5 and 3x + 2y = 11. (M'19)

Step 1: Condition for no solution
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 2x - ty = 5: a₁ = 2, b₁ = -t, c₁ = 5
For 3x + 2y = 11: a₂ = 3, b₂ = 2, c₂ = 11

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ 2/3 = -t/2

Step 4: Solve for t
2/3 = -t/2
Cross-multiplying: 4 = -3t
t = -4/3

Step 5: Verify c₁/c₂
c₁/c₂ = 5/11
Since 2/3 ≠ 5/11, the condition is satisfied.

∴ t = -4/3

The solution of the linear equation x + y = 5 are (1, 4), (2, 3) and (3, 2). The solution of another linear equation x - y = 1 are (3, 2), (2, 1) and (5, 4). Plot these points on a graph sheet and draw lines. (May 2022)

Step 1: Identify the equations
Equation 1: x + y = 5
Equation 2: x - y = 1

Step 2: Find the intersection point
Solving the equations:
x + y = 5 ...(1)
x - y = 1 ...(2)
Adding (1) and (2): 2x = 6 ⇒ x = 3
Substituting in (1): 3 + y = 5 ⇒ y = 2

Step 3: Graph description

Step 4: Interpretation
The lines intersect at (3, 2), which is the common solution to both equations.

∴ The lines intersect at (3, 2)

The solutions of the linear equation x + y = 8 are (1, 8), (2, 6) and (3, 5). The solutions of another linear equation 3x + 3y = 12 are (1, 3), (3, 1) and (9, 4). Plot these points on a graph sheet and draw lines.

Step 1: Simplify the equations
Equation 1: x + y = 8
Equation 2: 3x + 3y = 12 ⇒ Divide by 3: x + y = 4

Step 2: Analyze the equations
Both equations have the same left-hand side (x + y) but different right-hand sides (8 and 4).
This means the lines are parallel.

Step 3: Graph description

Step 4: Interpretation
The lines are parallel and will never intersect, so the system has no solution.

∴ The lines are parallel and the system has no solution

If the pair of linear equations 6x - 4y + 10 = 0 and 3x + ky + 6 = 0 represents parallel lines graphically, then find the value of 'k'. (Jun'23)

Step 1: Condition for parallel lines
For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 6x - 4y + 10 = 0: a₁ = 6, b₁ = -4, c₁ = 10
For 3x + ky + 6 = 0: a₂ = 3, b₂ = k, c₂ = 6

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ 6/3 = -4/k

Step 4: Solve for k
2 = -4/k
k = -4/2 = -2

Step 5: Verify c₁/c₂
c₁/c₂ = 10/6 = 5/3
Since 2 ≠ 5/3, the condition is satisfied.

∴ k = -2

Linear Equations - 2 Mark Questions

If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same. Is it true? If so, justify with an example. (M'15)

Step 1: Understanding the concept
Yes, this statement is true. When we multiply or divide both sides of a linear equation by the same non-zero number, we are performing an equivalent transformation that doesn't change the solution set.

Step 2: Example to justify
Consider the equation: 2x + 4 = 10
Solution: 2x = 6 ⇒ x = 3

Step 3: Multiply both sides by 2
(2x + 4) × 2 = 10 × 2 ⇒ 4x + 8 = 20
Solution: 4x = 12 ⇒ x = 3

Step 4: Divide both sides by 2
(2x + 4) ÷ 2 = 10 ÷ 2 ⇒ x + 2 = 5
Solution: x = 3

∴ Yes, the statement is true. Multiplying or dividing both sides by a non-zero number doesn't change the solution.

If the present ages of A and B are in ratio of 9 : 4 and after 7 years the ratio of the ages will be 5 : 3 then find their present ages. (J'15)

Step 1: Set up variables
Let present age of A = 9x years
Let present age of B = 4x years

Step 2: Set up equation for after 7 years
After 7 years:
Age of A = 9x + 7
Age of B = 4x + 7
Ratio = (9x + 7) : (4x + 7) = 5 : 3

Step 3: Form the equation
(9x + 7)/(4x + 7) = 5/3
Cross-multiplying: 3(9x + 7) = 5(4x + 7)

Step 4: Solve for x
27x + 21 = 20x + 35
27x - 20x = 35 - 21
7x = 14 ⇒ x = 2

Step 5: Find present ages
Age of A = 9 × 2 = 18 years
Age of B = 4 × 2 = 8 years

∴ Present ages: A = 18 years, B = 8 years

Solve the following pair of linear equations by substitution method: 2x - 3y = 19 and 3x - 2y = 21 (M'16)

Substitution Method:

Step 1: Express one variable in terms of the other
From first equation: 2x - 3y = 19 ⇒ 2x = 19 + 3y ⇒ x = (19 + 3y)/2

Step 2: Substitute in second equation
3x - 2y = 21
3[(19 + 3y)/2] - 2y = 21
(57 + 9y)/2 - 2y = 21

Step 3: Solve for y
Multiply throughout by 2: 57 + 9y - 4y = 42
57 + 5y = 42
5y = 42 - 57 = -15
y = -3

Step 4: Find x
x = (19 + 3y)/2 = (19 + 3(-3))/2 = (19 - 9)/2 = 10/2 = 5

∴ x = 5, y = -3

If the measure of angles of a triangle are x°, y° and 40°, and difference between the measures of angles x° and y° is 30°, then find values of x° and y°. (J'16)

Step 1: Use angle sum property of triangle
Sum of angles of a triangle = 180°
So, x + y + 40 = 180
x + y = 140 ...(1)

Step 2: Use the given difference
x - y = 30 ...(2) [Assuming x > y]

Step 3: Solve the equations
Adding (1) and (2):
(x + y) + (x - y) = 140 + 30
2x = 170 ⇒ x = 85

Step 4: Find y
From equation (1): 85 + y = 140 ⇒ y = 55

∴ x = 85°, y = 55°

Given the linear equation 3x + 4y = 11, write linear equations in two variables such that their geometrical representations form parallel lines and intersecting lines. (M'18)

Step 1: For parallel lines
For parallel lines, the ratios of coefficients of x and y should be equal but the constant term should be different.
Example: 3x + 4y = 15 (same coefficients for x and y, different constant)

Step 2: For intersecting lines
For intersecting lines, the ratios of coefficients of x and y should not be equal.
Example: 2x + 5y = 12 (different ratio of coefficients)

∴ Parallel lines: 3x + 4y = 15
Intersecting lines: 2x + 5y = 12

Solve the pair of linear equations 2x + 3y = 8 and x + 2y = 5 by Elimination method. (M'19)

Elimination Method:

Step 1: Make coefficients of one variable equal
Equations: 2x + 3y = 8 ...(1)
x + 2y = 5 ...(2)
Multiply equation (2) by 2: 2x + 4y = 10 ...(3)

Step 2: Eliminate x
Subtract equation (1) from equation (3):
(2x + 4y) - (2x + 3y) = 10 - 8
y = 2

Step 3: Find x
Substitute y = 2 in equation (2):
x + 2(2) = 5 ⇒ x + 4 = 5 ⇒ x = 1

∴ x = 1, y = 2

For what values of m the following system of equations will have no solution? Why?
mx + 4y = 10 and 9x + 12y = 30 (J'19)

Step 1: Condition for no solution
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For mx + 4y = 10: a₁ = m, b₁ = 4, c₁ = 10
For 9x + 12y = 30: a₂ = 9, b₂ = 12, c₂ = 30

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ m/9 = 4/12 = 1/3
m/9 = 1/3 ⇒ m = 3

Step 4: Check c₁/c₂
c₁/c₂ = 10/30 = 1/3
Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3, the lines are coincident, not parallel.

Step 5: Conclusion
For no solution, we need a₁/a₂ = b₁/b₂ ≠ c₁/c₂
But here when m = 3, a₁/a₂ = b₁/b₂ = c₁/c₂
So there is no value of m for which the system has no solution.

∴ There is no value of m for which the system has no solution.

Solve 2x + y = 5 and 5x + 3y = 11. (May 2022)

Step 1: Use substitution method
From first equation: 2x + y = 5 ⇒ y = 5 - 2x

Step 2: Substitute in second equation
5x + 3y = 11
5x + 3(5 - 2x) = 11
5x + 15 - 6x = 11

Step 3: Solve for x
-x + 15 = 11
-x = 11 - 15 = -4
x = 4

Step 4: Find y
y = 5 - 2x = 5 - 2(4) = 5 - 8 = -3

∴ x = 4, y = -3

Solve 3x + 2y = 11 and 2x + 3y = 4. (Aug 2022)

Elimination Method:

Step 1: Make coefficients of y equal
Equations: 3x + 2y = 11 ...(1)
2x + 3y = 4 ...(2)
Multiply (1) by 3: 9x + 6y = 33 ...(3)
Multiply (2) by 2: 4x + 6y = 8 ...(4)

Step 2: Eliminate y
Subtract (4) from (3):
(9x + 6y) - (4x + 6y) = 33 - 8
5x = 25 ⇒ x = 5

Step 3: Find y
Substitute x = 5 in equation (1):
3(5) + 2y = 11 ⇒ 15 + 2y = 11 ⇒ 2y = -4 ⇒ y = -2

∴ x = 5, y = -2

Linear Equations -4 Marks Solutions

Linear Equations - 4 Marks Solutions

Draw a graph for the following pair of linear equations in two variables and find their solution from the graph: 2x + y = 5 and 3x – 2y = 4 (M'15)

Graphical Method:

Step 1: Find points for 2x + y = 5

When x = 0: 2(0) + y = 5 → 0 + y = 5 → y = 5 → Point: (0, 5)

When x = 1: 2(1) + y = 5 → 2 + y = 5 → y = 3 → Point: (1, 3)

When x = 2: 2(2) + y = 5 → 4 + y = 5 → y = 1 → Point: (2, 1)

Step 2: Find points for 3x - 2y = 4

When x = 0: 3(0) - 2y = 4 → 0 - 2y = 4 → -2y = 4 → y = -2 → Point: (0, -2)

When x = 2: 3(2) - 2y = 4 → 6 - 2y = 4 → -2y = -2 → y = 1 → Point: (2, 1)

When x = 4: 3(4) - 2y = 4 → 12 - 2y = 4 → -2y = -8 → y = 4 → Point: (4, 4)

∴ The lines intersect at point (2, 1), so the solution is x = 2, y = 1

Draw the graphs of the following equations 3x – y – 2 = 0 and 2x + y – 8 = 0 on the graph paper.
i) Write down the co-ordinates of the point of intersection of the equations.
ii) Find the area of the triangle formed by the lines and the X-axis. (J'15)

Step 1: Find points for 3x - y - 2 = 0

When x = 0: 3(0) - y - 2 = 0 → 0 - y - 2 = 0 → -y = 2 → y = -2 → Point: (0, -2)

When x = 1: 3(1) - y - 2 = 0 → 3 - y - 2 = 0 → 1 - y = 0 → y = 1 → Point: (1, 1)

When x = 2: 3(2) - y - 2 = 0 → 6 - y - 2 = 0 → 4 - y = 0 → y = 4 → Point: (2, 4)

Step 2: Find points for 2x + y - 8 = 0

When x = 0: 2(0) + y - 8 = 0 → 0 + y - 8 = 0 → y = 8 → Point: (0, 8)

When x = 2: 2(2) + y - 8 = 0 → 4 + y - 8 = 0 → y - 4 = 0 → y = 4 → Point: (2, 4)

When x = 4: 2(4) + y - 8 = 0 → 8 + y - 8 = 0 → y = 0 → Point: (4, 0)

Step 3: Find intersection point

The lines intersect at (2, 4)

Step 4: Find area of triangle with X-axis

For 3x - y - 2 = 0: When y = 0, 3x - 0 - 2 = 0 → 3x = 2 → x = 2/3 ≈ 0.67

For 2x + y - 8 = 0: When y = 0, 2x + 0 - 8 = 0 → 2x = 8 → x = 4

Base = 4 - 2/3 = 10/3

Height = y-coordinate of intersection point = 4

Area = 1/2 × base × height = 1/2 × (10/3) × 4 = 20/3 ≈ 6.67 sq units

∴ i) Intersection point: (2, 4)
ii) Area of triangle: 20/3 sq units ≈ 6.67 sq units

Draw the graph for the equations 2x – 3y = 5 and 4x – 6y = 15 on the graph paper and check whether they are consistent or not. (J'15)

Step 1: Find points for 2x - 3y = 5

When x = 1: 2(1) - 3y = 5 → 2 - 3y = 5 → -3y = 3 → y = -1 → Point: (1, -1)

When x = 4: 2(4) - 3y = 5 → 8 - 3y = 5 → -3y = -3 → y = 1 → Point: (4, 1)

When x = 7: 2(7) - 3y = 5 → 14 - 3y = 5 → -3y = -9 → y = 3 → Point: (7, 3)

Step 2: Find points for 4x - 6y = 15

When x = 3: 4(3) - 6y = 15 → 12 - 6y = 15 → -6y = 3 → y = -0.5 → Point: (3, -0.5)

When x = 6: 4(6) - 6y = 15 → 24 - 6y = 15 → -6y = -9 → y = 1.5 → Point: (6, 1.5)

When x = 9: 4(9) - 6y = 15 → 36 - 6y = 15 → -6y = -21 → y = 3.5 → Point: (9, 3.5)

Step 3: Check consistency

For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, then lines are parallel and inconsistent

Here: 2/4 = 1/2, -3/-6 = 1/2, 5/15 = 1/3

Since 1/2 = 1/2 ≠ 1/3, the lines are parallel

∴ The equations are inconsistent (no solution) as the lines are parallel

Draw the graph for the following pair of linear equations in two variables and find their solution from the graph. 3x– 2y = 2 and 2x + y = 6 (M'16)

Step 1: Find points for 3x - 2y = 2

When x = 0: 3(0) - 2y = 2 → 0 - 2y = 2 → -2y = 2 → y = -1 → Point: (0, -1)

When x = 2: 3(2) - 2y = 2 → 6 - 2y = 2 → -2y = -4 → y = 2 → Point: (2, 2)

When x = 4: 3(4) - 2y = 2 → 12 - 2y = 2 → -2y = -10 → y = 5 → Point: (4, 5)

Step 2: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 2: 2(2) + y = 6 → 4 + y = 6 → y = 2 → Point: (2, 2)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

∴ The lines intersect at point (2, 2), so the solution is x = 2, y = 2

Draw the graph for the equations 2x – y – 4 = 0 and x + y = 0 on the graph paper and check whether they are consistent or not. (J'16)

Step 1: Find points for 2x - y - 4 = 0

When x = 0: 2(0) - y - 4 = 0 → 0 - y - 4 = 0 → -y = 4 → y = -4 → Point: (0, -4)

When x = 2: 2(2) - y - 4 = 0 → 4 - y - 4 = 0 → -y = 0 → y = 0 → Point: (2, 0)

When x = 4: 2(4) - y - 4 = 0 → 8 - y - 4 = 0 → 4 - y = 0 → y = 4 → Point: (4, 4)

Step 2: Find points for x + y = 0

When x = -2: -2 + y = 0 → y = 2 → Point: (-2, 2)

When x = 0: 0 + y = 0 → y = 0 → Point: (0, 0)

When x = 2: 2 + y = 0 → y = -2 → Point: (2, -2)

Step 3: Find intersection point algebraically

From x + y = 0, we have y = -x

Substitute in 2x - y - 4 = 0: 2x - (-x) - 4 = 0 → 2x + x - 4 = 0 → 3x = 4 → x = 4/3

Then y = -4/3

∴ The equations are consistent (have a unique solution) and intersect at (4/3, -4/3)

Draw the graph of 2x + y = 6 and 2x – y + 2 = 0 and find the solution from the graph. (M'17)

Step 1: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 1: 2(1) + y = 6 → 2 + y = 6 → y = 4 → Point: (1, 4)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

Step 2: Find points for 2x - y + 2 = 0

When x = 0: 2(0) - y + 2 = 0 → 0 - y + 2 = 0 → -y = -2 → y = 2 → Point: (0, 2)

When x = 1: 2(1) - y + 2 = 0 → 2 - y + 2 = 0 → 4 - y = 0 → y = 4 → Point: (1, 4)

When x = 2: 2(2) - y + 2 = 0 → 4 - y + 2 = 0 → 6 - y = 0 → y = 6 → Point: (2, 6)

∴ The lines intersect at point (1, 4), so the solution is x = 1, y = 4

Show that the following pair of equations are consistent and show them graphically: x + 3y = 6 and 2x– 3y = 12 (J'17)

Step 1: Check consistency

For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

If a₁/a₂ ≠ b₁/b₂, then lines intersect and are consistent

Here: 1/2 = 0.5, 3/-3 = -1

Since 0.5 ≠ -1, the lines intersect and are consistent

Step 2: Find points for x + 3y = 6

When x = 0: 0 + 3y = 6 → 3y = 6 → y = 2 → Point: (0, 2)

When x = 3: 3 + 3y = 6 → 3y = 3 → y = 1 → Point: (3, 1)

When x = 6: 6 + 3y = 6 → 3y = 0 → y = 0 → Point: (6, 0)

Step 3: Find points for 2x - 3y = 12

When x = 0: 2(0) - 3y = 12 → 0 - 3y = 12 → -3y = 12 → y = -4 → Point: (0, -4)

When x = 3: 2(3) - 3y = 12 → 6 - 3y = 12 → -3y = 6 → y = -2 → Point: (3, -2)

When x = 6: 2(6) - 3y = 12 → 12 - 3y = 12 → -3y = 0 → y = 0 → Point: (6, 0)

∴ The equations are consistent (have a unique solution) and intersect at (6, 0)

Solve the following pair of linear equations by graph method. 2x + y = 6 and 2x – y + 2 = 0. (J'18)

Step 1: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 1: 2(1) + y = 6 → 2 + y = 6 → y = 4 → Point: (1, 4)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

Step 2: Find points for 2x - y + 2 = 0

When x = 0: 2(0) - y + 2 = 0 → 0 - y + 2 = 0 → -y = -2 → y = 2 → Point: (0, 2)

When x = 1: 2(1) - y + 2 = 0 → 2 - y + 2 = 0 → 4 - y = 0 → y = 4 → Point: (1, 4)

When x = 2: 2(2) - y + 2 = 0 → 4 - y + 2 = 0 → 6 - y = 0 → y = 6 → Point: (2, 6)

∴ The lines intersect at point (1, 4), so the solution is x = 1, y = 4

Solve the equations by graphically 3x + 4y = 10 and 4x – 3y = 5. (M'19)

Step 1: Find points for 3x + 4y = 10

When x = 0: 3(0) + 4y = 10 → 0 + 4y = 10 → 4y = 10 → y = 2.5 → Point: (0, 2.5)

When x = 2: 3(2) + 4y = 10 → 6 + 4y = 10 → 4y = 4 → y = 1 → Point: (2, 1)

When x = 4: 3(4) + 4y = 10 → 12 + 4y = 10 → 4y = -2 → y = -0.5 → Point: (4, -0.5)

Step 2: Find points for 4x - 3y = 5

When x = 0: 4(0) - 3y = 5 → 0 - 3y = 5 → -3y = 5 → y = -5/3 ≈ -1.67 → Point: (0, -1.67)

When x = 2: 4(2) - 3y = 5 → 8 - 3y = 5 → -3y = -3 → y = 1 → Point: (2, 1)

When x = 4: 4(4) - 3y = 5 → 16 - 3y = 5 → -3y = -11 → y = 11/3 ≈ 3.67 → Point: (4, 3.67)

∴ The lines intersect at point (2, 1), so the solution is x = 2, y = 1

Sum of the present ages of two friends are 23 years, five years ago product of their ages was 42. Find their ages 5 years hence. (M'19)

Step 1: Set up equations

Let present ages be x and y years

x + y = 23 ...(1)

Five years ago: (x-5)(y-5) = 42 ...(2)

Step 2: Expand equation (2)

(x-5)(y-5) = 42

xy - 5x - 5y + 25 = 42

xy - 5(x+y) = 17

From (1): x+y = 23, so

xy - 5(23) = 17

xy - 115 = 17

xy = 132

Step 3: Solve the system

We have: x + y = 23 and xy = 132

This forms a quadratic: t² - 23t + 132 = 0

Discriminant = 23² - 4(1)(132) = 529 - 528 = 1

t = [23 ± √1]/2 = [23 ± 1]/2

t = 12 or t = 11

Step 4: Find ages 5 years hence

Present ages: 12 and 11 years

Ages 5 years hence: 17 and 16 years

∴ Their ages 5 years hence will be 17 years and 16 years

Draw the graph of x + y = 11 and x - y = 5. Find the solution of the pair of linear equations (J'19)

Step 1: Find points for x + y = 11

When x = 0: 0 + y = 11 → y = 11 → Point: (0, 11)

When x = 5: 5 + y = 11 → y = 6 → Point: (5, 6)

When x = 11: 11 + y = 11 → y = 0 → Point: (11, 0)

Step 2: Find points for x - y = 5

When x = 0: 0 - y = 5 → -y = 5 → y = -5 → Point: (0, -5)

When x = 5: 5 - y = 5 → -y = 0 → y = 0 → Point: (5, 0)

When x = 10: 10 - y = 5 → -y = -5 → y = 5 → Point: (10, 5)

∴ The lines intersect at point (8, 3), so the solution is x = 8, y = 3

Author

zaheerpashamd

M.Sc,B.Ed: SDVR ZPHS B.Gangaram Sathupally(md) Khammam(dt)
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