RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables More Ques

Multiple Choice Questions

 

Question 1. If y = 2x - 3 and y = 5 then value of x is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4
In simple words: We are given two equations involving x and y. Since both equations tell us what y is, we can set them equal to each other to find x. Substitute y=5 into the first equation to solve for x.

🎯 Exam Tip: For these types of problems, substitute the known value into the equation and then solve for the unknown variable. Always double-check your answer by putting the calculated x value back into the original equations.

 

Question 2. If 2x + y = 6 then ordered pair satisfying this equation is:
(a) (1, 2)
(b) (2, 1)
(c) (2, 2)
(d) (1, 1)
Answer: (c) (2, 2)
In simple words: To find which pair works, put the x and y values from each option into the equation. The pair that makes both sides of the equation equal is the correct answer. The coordinates (2,2) fit the equation, meaning when x=2 and y=2, 2(2)+2 = 6.

🎯 Exam Tip: The easiest way to solve such an MCQ is to plug in the given options into the equation. The option that satisfies the equation is the correct answer.

 

Question 3. If \( \frac {4}{x} + 5y = 7 \) and \( x = \frac {-4}{3} \) then value of y is:
(a) \( \frac {37}{15} \)
(b) 2
(c) \( \frac {1}{2} \)
(d) \( \frac {1}{3} \)
Answer: (b) 2
In simple words: We know the value of x, so we replace x in the first equation with \( \frac{-4}{3} \). Then, we solve the new equation to find the value of y. Plugging in x makes the equation easier to solve.

🎯 Exam Tip: Always pay attention to negative signs and fractions when substituting values. Use proper order of operations (PEMDAS/BODMAS) for accurate calculations.

 

Question 4. If \( \frac {3}{x} + 4y = 5 \) and y = 1 then value of x is:
(a) 3
(b) \( \frac {1}{3} \)
(c) 3
(d) -3
Answer: (a) 3
In simple words: We are given the value of y. Put y=1 into the equation \( \frac{3}{x} + 4y = 5 \). After that, solve the equation to find the value of x.

🎯 Exam Tip: When dealing with variables in the denominator, remember that the variable cannot be zero. Cross-multiplication can simplify the equation once you isolate the term with x.

 

Question 6. If y and x are the digits in units place and tens place respectively then the number is:
(a) 10x + y
(b) 10y + x
(c) x + y
(d) xy
Answer: (a) 10x + y
In simple words: In a two-digit number, the tens place digit is multiplied by 10 and then added to the units place digit. For example, if x is 2 and y is 3, the number is 23, which is (10 * 2) + 3.

🎯 Exam Tip: Remember the place value system: the digit in the tens place has a value ten times its face value, while the digit in the units place has its face value.

 

Question 7. A boy is now one third as old as his mother. What will be the age of the boy after 12 years if mother's present age is x?
(a) \( \frac {x}{3}+12 \)
(b) \( \frac {x+12}{2} \)
(c) x + 4
(d) \( \frac {x}{3} - 12 \)
Answer: (a) \( \frac {x}{3}+12 \)
In simple words: First, find the boy's current age by dividing the mother's age (x) by three. Then, add 12 years to that result to find his age after 12 years. Always write ages as positive values.

🎯 Exam Tip: Break down word problems into smaller steps: first determine the current situation, then apply changes over time. Be careful with 'of' (multiplication) and 'more than' (addition).

 

Question 8. The point on the x axis is:
(a) (2, 3)
(b) (2, 0)
(c) (0, 2)
(d) (2, 2)
Answer: (b) (2, 0)
In simple words: Any point that lies on the x-axis will always have its y-coordinate as zero. The x-coordinate can be any real number. So, the point (2, 0) is on the x-axis.

🎯 Exam Tip: Points on the x-axis have the form (x, 0), and points on the y-axis have the form (0, y). This is a basic concept in coordinate geometry.

 

Question 9. Co-ordinates of the origin are:
(a) (0, 0)
(b) (0, 1)
(c) (1,0)
(d) (1, 1)
Answer: (a) (0, 0)
In simple words: The origin is the central point in a coordinate system where the x-axis and y-axis cross each other. Its coordinates are always zero for both x and y.

🎯 Exam Tip: The origin (0, 0) is the reference point for all other coordinates in a Cartesian system. It represents the starting point for measuring distances along the axes.

 

Question 11. Express y in terms of x in \( 5y – 3x – 10 = 0 \). Also find the point where the line \( 5y – 3x - 10 = 0 \) will intersect the y-axis.
Answer: The given equation is \( 5y - 3x - 10 = 0 \).
To express y in terms of x, we rearrange the equation:
\( 5y = 3x + 10 \)
Now, divide by 5:
\( \implies y = \frac {3}{5}x + 2 \)
The line will intersect the y-axis when the x-coordinate is 0. This is because all points on the y-axis have an x-value of zero.
Substitute \( x = 0 \) into the equation for y:
\( y = \frac {3}{5}(0) + 2 \)
\( \implies y = 0 + 2 \)
\( \implies y = 2 \)
So, the coordinates of the point where the line intersects the y-axis are (0, 2).
In simple words: First, rewrite the equation to get 'y' by itself on one side, which shows how y changes with x. Then, to find where the line crosses the y-axis, set x to zero and solve for y.

🎯 Exam Tip: To find the y-intercept, set x = 0 in the equation. To find the x-intercept, set y = 0 in the equation. This is a fundamental concept for graphing linear equations.

 

Question 12. By taking the values of x = -2 to x = 2 and also included value of x for the equation y = 2x + 1. Form the table and draw the graph of the equation.
Answer: We need to create a table of values for the equation \( y = 2x + 1 \) by using x values from -2 to 2.
When \( x = -2 \): \( y = 2(-2) + 1 = -4 + 1 = -3 \)
When \( x = -1 \): \( y = 2(-1) + 1 = -2 + 1 = -1 \)
When \( x = 0 \): \( y = 2(0) + 1 = 0 + 1 = 1 \)
When \( x = 1 \): \( y = 2(1) + 1 = 2 + 1 = 3 \)
When \( x = 2 \): \( y = 2(2) + 1 = 4 + 1 = 5 \)
Now we form the table with these values:

To draw the graph, plot these ordered pairs (x, y) on a coordinate plane and connect them with a straight line. This visual representation helps understand the relationship between x and y.
In simple words: For the given equation, put in different numbers for x to find the matching y values. Write these pairs in a table. Then, draw these points on a graph and connect them with a straight line.

🎯 Exam Tip: When drawing graphs of linear equations, it's good practice to calculate at least three points to ensure accuracy, as two points define a line and a third acts as a check.

 

Question 13. Solve the following simultaneous equations:
\( 0.5x + 0.6y = 2.3 \)
\( 0.2x + 0.7y = 2.3 \)
Answer: The given equations are:
\( 0.5x + 0.6y = 2.3 \) ...(i)
\( 0.2x + 0.7y = 2.3 \) ...(ii)
To eliminate one variable, we can multiply the equations to make the coefficients of x (or y) the same.
Multiply equation (i) by 2 and equation (ii) by 5:
\( 2 \times (0.5x + 0.6y) = 2 \times 2.3 \implies 1.0x + 1.2y = 4.6 \) ...(iii)
\( 5 \times (0.2x + 0.7y) = 5 \times 2.3 \implies 1.0x + 3.5y = 11.5 \) ...(iv)
Now, subtract equation (iii) from equation (iv) to eliminate x:
\( (1.0x + 3.5y) - (1.0x + 1.2y) = 11.5 - 4.6 \)
\( \implies 2.3y = 6.9 \)
\( \implies y = \frac{6.9}{2.3} \)
\( \implies y = 3 \)
Substitute the value of \( y = 3 \) into equation (i):
\( 0.5x + 0.6(3) = 2.3 \)
\( \implies 0.5x + 1.8 = 2.3 \)
\( \implies 0.5x = 2.3 - 1.8 \)
\( \implies 0.5x = 0.5 \)
\( \implies x = \frac{0.5}{0.5} \)
\( \implies x = 1 \)
Thus, the required solution is \( x = 1, y = 3 \). Finding the common solution for both equations means finding the point where their graphs intersect.
In simple words: We have two math problems with x and y. To solve them, we change the equations so that we can remove either x or y. Then, we find the value of the remaining letter, and use that to find the value of the first letter.

🎯 Exam Tip: When dealing with decimal coefficients, you can multiply the entire equation by a power of 10 to convert them into integers, making calculations simpler and reducing error possibilities.

 

Question 14. Solve the following system of equations:
\( 2x + 3y = 9 \)
\( 3x + 4y = 5 \)
Answer: The given equations are:
\( 2x + 3y = 9 \) ...(i)
\( 3x + 4y = 5 \) ...(ii)
To solve this system, we can use the elimination method. We will multiply equation (i) by 3 and equation (ii) by 2 to make the coefficients of x equal.
Multiply equation (i) by 3: \( 3 \times (2x + 3y) = 3 \times 9 \implies 6x + 9y = 27 \) ...(iii)
Multiply equation (ii) by 2: \( 2 \times (3x + 4y) = 2 \times 5 \implies 6x + 8y = 10 \) ...(iv)
Now, subtract equation (iv) from equation (iii):
\( (6x + 9y) - (6x + 8y) = 27 - 10 \)
\( \implies y = 17 \)
Next, substitute the value of \( y = 17 \) into equation (i) to find x:
\( 2x + 3(17) = 9 \)
\( \implies 2x + 51 = 9 \)
\( \implies 2x = 9 - 51 \)
\( \implies 2x = -42 \)
\( \implies x = \frac{-42}{2} \)
\( \implies x = -21 \)
So, the solution to this system of equations is \( x = -21 \) and \( y = 17 \). These values satisfy both equations simultaneously.
In simple words: We have two math problems with two unknowns. We make one of the unknowns have the same number in both problems by multiplying. Then we subtract one problem from the other to remove that unknown. After that, we find the other unknown and put its value back to find the first one.

🎯 Exam Tip: The elimination method is often efficient for solving systems of equations, especially when coefficients can be easily made equal. Always check your final answers by substituting them back into both original equations.

 

Question 15. Solve the following system of equations:
\( \frac {1}{2x} - \frac {1}{y} = -1 \)
\( \frac {1}{x} + \frac {1}{2y} = 8 \)
where \( x \neq 0, y \neq 0 \)
Answer: The given equations are:
\( \frac {1}{2x} - \frac {1}{y} = -1 \) ...(i)
\( \frac {1}{x} + \frac {1}{2y} = 8 \) ...(ii)
This type of equation can be simplified by substituting new variables. Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \).
Substituting these into the equations, we get:
From (i): \( \frac{1}{2}a - b = -1 \) ...(iii)
From (ii): \( a + \frac{1}{2}b = 8 \) ...(iv)
Now we have a simpler system of linear equations in 'a' and 'b'.
Multiply equation (iii) by 2 to clear the fraction:
\( 2 \times (\frac{1}{2}a - b) = 2 \times (-1) \implies a - 2b = -2 \) ...(v)
Multiply equation (iv) by 2 to clear the fraction:
\( 2 \times (a + \frac{1}{2}b) = 2 \times 8 \implies 2a + b = 16 \) ...(vi)
From equation (v), we can express 'a' in terms of 'b':
\( a = 2b - 2 \)
Substitute this expression for 'a' into equation (vi):
\( 2(2b - 2) + b = 16 \)
\( \implies 4b - 4 + b = 16 \)
\( \implies 5b - 4 = 16 \)
\( \implies 5b = 16 + 4 \)
\( \implies 5b = 20 \)
\( \implies b = \frac{20}{5} \)
\( \implies b = 4 \)
Now substitute \( b = 4 \) back into the expression for 'a':
\( a = 2(4) - 2 \)
\( \implies a = 8 - 2 \)
\( \implies a = 6 \)
Finally, we convert back to x and y using our original substitutions:
Since \( a = \frac{1}{x} \), then \( 6 = \frac{1}{x} \implies x = \frac{1}{6} \)
Since \( b = \frac{1}{y} \), then \( 4 = \frac{1}{y} \implies y = \frac{1}{4} \)
The solution to the system is \( x = \frac{1}{6} \) and \( y = \frac{1}{4} \). This method makes complex equations easier to manage.
In simple words: First, we change the complex fractions into simpler letters like 'a' and 'b'. Then we solve for these new letters using common methods. After finding 'a' and 'b', we change them back to find the original x and y values.

🎯 Exam Tip: When an equation involves variables in the denominator, simplify it by substituting new variables for the reciprocal terms (e.g., \( \frac{1}{x} \)). This transforms non-linear equations into linear ones, which are easier to solve.

 

Question 16. Two numbers are such that if 7 is added to the smaller number, then the sum is twice the greater and if 4 is added to the greater number, then the sum is thrice the smaller number. Find the two numbers.
Answer: Let the two numbers be x and y. Assume that \( x \) is the greater number and \( y \) is the smaller number, so \( x > y \).
From the first condition: "if 7 is added to the smaller number, then the sum is twice the greater"
\( y + 7 = 2x \)
Rearranging this, we get:
\( 2x - y = 7 \) ...(i)
From the second condition: "if 4 is added to the greater number, then the sum is thrice the smaller number"
\( x + 4 = 3y \)
Rearranging this, we get:
\( x - 3y = -4 \) ...(ii)
Now we have a system of two linear equations:
1. \( 2x - y = 7 \)
2. \( x - 3y = -4 \)
To solve this, we can multiply equation (ii) by 2 to make the x-coefficients match:
\( 2 \times (x - 3y) = 2 \times (-4) \implies 2x - 6y = -8 \) ...(iii)
Now, subtract equation (iii) from equation (i):
\( (2x - y) - (2x - 6y) = 7 - (-8) \)
\( \implies 2x - y - 2x + 6y = 7 + 8 \)
\( \implies 5y = 15 \)
\( \implies y = \frac{15}{5} \)
\( \implies y = 3 \)
Substitute \( y = 3 \) back into equation (ii) to find x:
\( x - 3(3) = -4 \)
\( \implies x - 9 = -4 \)
\( \implies x = -4 + 9 \)
\( \implies x = 5 \)
So, the two numbers are 5 and 3. We check our assumption: 5 is indeed greater than 3, so our assignment of x and y was correct.
In simple words: We first set up two math problems using letters for the unknown numbers, based on the story. Then we solve these two problems together to find the values of both numbers.

🎯 Exam Tip: Always define your variables clearly (e.g., let x be the greater number, y be the smaller number). Translate each sentence of the word problem into a separate mathematical equation. This helps avoid confusion.

 

Question 17. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Answer: Let the fraction be \( \frac{x}{y} \), where x is the numerator and y is the denominator.
From the first condition: "The numerator of a fraction is 4 less than the denominator."
\( x = y - 4 \)
Rearranging this, we get:
\( x - y + 4 = 0 \) ...(i)
From the second condition: "If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator."
New numerator = \( x - 2 \)
New denominator = \( y + 1 \)
So, \( y + 1 = 8(x - 2) \)
\( \implies y + 1 = 8x - 16 \)
Rearranging this equation, we get:
\( 8x - y - 17 = 0 \) ...(ii)
Now we have a system of two linear equations:
1. \( x - y + 4 = 0 \implies x - y = -4 \)
2. \( 8x - y - 17 = 0 \implies 8x - y = 17 \)
To solve this, subtract equation (i) from equation (ii):
\( (8x - y) - (x - y) = 17 - (-4) \)
\( \implies 8x - y - x + y = 17 + 4 \)
\( \implies 7x = 21 \)
\( \implies x = \frac{21}{7} \)
\( \implies x = 3 \)
Substitute \( x = 3 \) back into equation (i):
\( 3 - y = -4 \)
\( \implies -y = -4 - 3 \)
\( \implies -y = -7 \)
\( \implies y = 7 \)
Thus, the original fraction is \( \frac{x}{y} = \frac{3}{7} \). Word problems often require careful conversion into algebraic expressions.
In simple words: We use letters for the top and bottom parts of the fraction. We write two math problems based on the story. Then we solve these problems to find what those letters stand for.

🎯 Exam Tip: Clearly identify the numerator and denominator and represent them with variables. Be precise when translating "less than," "decreased by," and "increased by" into algebraic expressions.

 

Question 18. 5 books and 7 pens together cost Rs 79, whereas 7 books and 5 pens together cost Rs 77. Find the total cost of 1 book and 2 pens.
Answer: Let the cost of 1 book be Rs x and the cost of 1 pen be Rs y.
From the first condition: "5 books and 7 pens together cost Rs 79"
\( 5x + 7y = 79 \) ...(i)
From the second condition: "7 books and 5 pens together cost Rs 77"
\( 7x + 5y = 77 \) ...(ii)
We have a system of two linear equations. We can solve this by elimination or substitution.
To eliminate y, multiply equation (i) by 5 and equation (ii) by 7:
\( 5 \times (5x + 7y) = 5 \times 79 \implies 25x + 35y = 395 \) ...(iii)
\( 7 \times (7x + 5y) = 7 \times 77 \implies 49x + 35y = 539 \) ...(iv)
Subtract equation (iii) from equation (iv):
\( (49x + 35y) - (25x + 35y) = 539 - 395 \)
\( \implies 49x - 25x = 144 \)
\( \implies 24x = 144 \)
\( \implies x = \frac{144}{24} \)
\( \implies x = 6 \)
Now substitute \( x = 6 \) into equation (i):
\( 5(6) + 7y = 79 \)
\( \implies 30 + 7y = 79 \)
\( \implies 7y = 79 - 30 \)
\( \implies 7y = 49 \)
\( \implies y = \frac{49}{7} \)
\( \implies y = 7 \)
So, the cost of 1 book (x) is Rs 6 and the cost of 1 pen (y) is Rs 7.
We need to find the total cost of 1 book and 2 pens:
Cost = \( 1 \times x + 2 \times y \)
Cost = \( 1 \times 6 + 2 \times 7 \)
Cost = \( 6 + 14 \)
Cost = Rs 20.
In simple words: We used letters to stand for the cost of one book and one pen. We made two math problems from the given information. We solved these to find the price of each item, then added up the cost for one book and two pens.

🎯 Exam Tip: Define variables for the price of each item clearly. After finding the individual prices, make sure to answer the specific question asked (e.g., total cost of 1 book and 2 pens), not just the individual prices.

 

Question 19. When a number consisting of two digits is multiplied by 9, it becomes equal to twice the number obtained by reversing the digits of the original number. If the difference of two digits of the number is 7, find the number.
Answer: Let the two-digit number be \( 10x + y \), where x is the tens place digit and y is the units place digit. Here, x must be from 1 to 9, and y must be from 0 to 9.
The number obtained by reversing the digits is \( 10y + x \).
From the first condition: "When a number consisting of two digits is multiplied by 9, it becomes equal to twice the number obtained by reversing the digits"
\( 9(10x + y) = 2(10y + x) \)
\( \implies 90x + 9y = 20y + 2x \)
Rearranging the terms:
\( \implies 90x - 2x = 20y - 9y \)
\( \implies 88x = 11y \)
Divide both sides by 11:
\( \implies 8x = y \) ...(i)
From the second condition: "If the difference of two digits of the number is 7"
There are two possibilities: \( x - y = 7 \) or \( y - x = 7 \).
If \( x - y = 7 \): Substitute \( y = 8x \) from (i) into this equation:
\( x - 8x = 7 \implies -7x = 7 \implies x = -1 \). This is not possible as a digit must be positive.
So, we must use the other possibility: \( y - x = 7 \) ...(ii)
Substitute \( y = 8x \) from (i) into equation (ii):
\( 8x - x = 7 \)
\( \implies 7x = 7 \)
\( \implies x = 1 \)
Now find y using \( y = 8x \):
\( y = 8(1) \implies y = 8 \)
So, the tens digit is 1 and the units digit is 8. The number is \( 10x + y = 10(1) + 8 = 18 \). We can check: \( 9 \times 18 = 162 \). Reversing digits gives 81, and \( 2 \times 81 = 162 \). Also, \( 8 - 1 = 7 \). All conditions are met.
In simple words: We represent the two-digit number using two letters for its digits. Then, we write two math problems based on the rules given in the story. Solving these problems helps us find the value of each digit and thus the original number.

🎯 Exam Tip: For two-digit number problems, remember to represent the number as \( 10 \times \text{tens digit} + \text{units digit} \). When reversing digits, swap the tens and units places accordingly. Always consider both possibilities for digit differences (\( x-y \) or \( y-x \)).

 

Question 20. In a \( \triangle ABC \), \( \angle A = x^\circ \), \( \angle B = 3x^\circ \) and \( \angle C = y^\circ \). If \( 5x – 3y + 30 = 0 \), then prove that it is a right angled triangle.
Answer: In \( \triangle ABC \), we are given the angles:
\( \angle A = x^\circ \)
\( \angle B = 3x^\circ \)
\( \angle C = y^\circ \)
We know that the sum of angles in a triangle is \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \implies x + 3x + y = 180 \)
\( \implies 4x + y = 180 \) ...(i)
We are also given another equation relating x and y:
\( 5x - 3y + 30 = 0 \)
\( \implies 5x - 3y = -30 \) ...(ii)
Now we have a system of two linear equations:
1. \( 4x + y = 180 \)
2. \( 5x - 3y = -30 \)
From equation (i), express y in terms of x:
\( y = 180 - 4x \)
Substitute this expression for y into equation (ii):
\( 5x - 3(180 - 4x) = -30 \)
\( \implies 5x - 540 + 12x = -30 \)
\( \implies 17x - 540 = -30 \)
\( \implies 17x = 540 - 30 \)
\( \implies 17x = 510 \)
\( \implies x = \frac{510}{17} \)
\( \implies x = 30^\circ \)
Now substitute \( x = 30^\circ \) back into \( y = 180 - 4x \):
\( y = 180 - 4(30) \)
\( \implies y = 180 - 120 \)
\( \implies y = 60^\circ \)
Now we find the values of the angles of the triangle:
\( \angle A = x^\circ = 30^\circ \)
\( \angle B = 3x^\circ = 3(30^\circ) = 90^\circ \)
\( \angle C = y^\circ = 60^\circ \)
Since one of the angles, \( \angle B \), is \( 90^\circ \), the triangle ABC is a right-angled triangle. This demonstrates how algebra can be used to prove geometric properties.
In simple words: We used the fact that all angles in a triangle add up to 180 degrees to make one equation. Then, we used the other given math problem. We solved these two math problems to find the values of all the angles. Since one angle came out to be 90 degrees, it proved the triangle is a right-angled triangle.

🎯 Exam Tip: Always remember the fundamental property that the sum of angles in a triangle is 180 degrees. If one angle calculates to 90 degrees, it automatically proves the triangle is right-angled.

 

Question 21. Solve the following pair of equations graphically.
(i) \( x + y = 4 \); \( x = y \)
(ii) \( x + y = 3 \); \( 2x + 5y = 12 \)
(iii) \( 2x – 3y - 6 = 0 \); \( 2x + y + 10 = 0 \)
(iv) \( 2x + y – 3 = 0 \); \( 2x – 3y - 7 = 0 \)
Answer:
(i) Given equations:
\( x + y = 4 \implies y = 4 - x \) ...(i-a)
\( x = y \) ...(i-b)
For \( y = 4 - x \):
When \( x = 1 \), \( y = 4 - 1 = 3 \)
When \( x = 2 \), \( y = 4 - 2 = 2 \)
When \( x = 3 \), \( y = 4 - 3 = 1 \)
When \( x = 4 \), \( y = 4 - 4 = 0 \)
Table for equation (i-a):

For \( y = x \):
When \( x = 0 \), \( y = 0 \)
When \( x = 1 \), \( y = 1 \)
When \( x = 2 \), \( y = 2 \)
When \( x = 3 \), \( y = 3 \)
Table for equation (i-b):

Plot the points (1, 3), (2, 2), (3, 1), (4, 0) for the first line and (0, 0), (1, 1), (2, 2), (3, 3) for the second line on a graph paper. The point where these two lines intersect is (2, 2). This intersection point is the solution to the system of equations.
Hence, the required solution is \( x = 2, y = 2 \).

(ii) Given equations:
\( x + y = 3 \implies y = 3 - x \) ...(ii-a)
\( 2x + 5y = 12 \implies y = \frac{12-2x}{5} \) ...(ii-b)
For \( y = 3 - x \):
When \( x = 0 \), \( y = 3 - 0 = 3 \)
When \( x = 3 \), \( y = 3 - 3 = 0 \)
When \( x = 2 \), \( y = 3 - 2 = 1 \)
Table for equation (ii-a):

For \( y = \frac{12-2x}{5} \):
When \( x = 1 \), \( y = \frac{12-2(1)}{5} = \frac{10}{5} = 2 \)
When \( x = 6 \), \( y = \frac{12-2(6)}{5} = \frac{0}{5} = 0 \)
When \( x = 11 \), \( y = \frac{12-2(11)}{5} = \frac{12-22}{5} = \frac{-10}{5} = -2 \)
Table for equation (ii-b):

Plot these points for both lines on a graph paper. The intersection point of these two lines will be the solution.
From the graph, we observe that the lines intersect at (1, 2). Thus, the solution is \( x = 1, y = 2 \).

(iii) Given equations:
\( 2x - 3y - 6 = 0 \implies 2x + y + 10 = 0 \)
From the first equation: \( 2x - 3y - 6 = 0 \implies 2x = 3y + 6 \implies x = \frac{3y+6}{2} \) ...(iii-a)
When \( y = 0 \), \( x = \frac{3(0)+6}{2} = \frac{6}{2} = 3 \)
When \( y = -1 \), \( x = \frac{3(-1)+6}{2} = \frac{3}{2} = 1.5 \)
When \( y = -2 \), \( x = \frac{3(-2)+6}{2} = \frac{0}{2} = 0 \)
Table for equation (iii-a):

From the second equation: \( 2x + y + 10 = 0 \implies y = -2x - 10 \) ...(iii-b)
When \( x = -2 \), \( y = -2(-2) - 10 = 4 - 10 = -6 \)
When \( x = -3 \), \( y = -2(-3) - 10 = 6 - 10 = -4 \)
When \( x = -4 \), \( y = -2(-4) - 10 = 8 - 10 = -2 \)
Table for equation (iii-b):

Plot these points for both lines on a graph paper. The intersection point of these two lines will be the solution.
From the graph, we find that the lines are intersecting at a point (-3, -4). Thus, the solution is \( x = -3, y = -4 \).

(iv) Given equations:
\( 2x + y - 3 = 0 \implies y = -2x + 3 \) ...(iv-a)
\( 2x - 3y - 7 = 0 \implies x = \frac{3y+7}{2} \) ...(iv-b)
For \( y = -2x + 3 \):
When \( x = 0 \), \( y = -2(0) + 3 = 3 \)
When \( x = 1 \), \( y = -2(1) + 3 = 1 \)
When \( x = 2 \), \( y = -2(2) + 3 = -1 \)
Table for equation (iv-a):

For \( x = \frac{3y+7}{2} \):
When \( y = 0 \), \( x = \frac{3(0)+7}{2} = \frac{7}{2} = 3.5 \)
When \( y = 1 \), \( x = \frac{3(1)+7}{2} = \frac{10}{2} = 5 \)
When \( y = -1 \), \( x = \frac{3(-1)+7}{2} = \frac{4}{2} = 2 \)
Table for equation (iv-b):

Plot these points for both lines on a graph paper. The intersection point of these two lines will be the solution.
From the graph, we find that the lines are intersecting at a point (2, -1). Thus, the solution is \( x = 2, y = -1 \).
In simple words: For each pair of equations, we make a table of matching x and y values for each line. Then, we draw both lines on a graph. The spot where the two lines cross over is the answer to the equations.

🎯 Exam Tip: When solving graphically, calculate at least three points for each line to ensure accuracy. Make sure your graph paper has a clear scale, and precisely mark the intersection point as your solution.

 

Question 22. Solve the following system of equations \( 2x - y = 1 \) and \( x + 2y = 8 \) graphically. Also, write the co-ordinate of the point where these lines intersect the y-axis.
Answer:
To solve these equations graphically, we first need to create tables of values for each equation and then plot those points. We will find points for two equations and then graph them on a coordinate plane. The point where the two lines cross each other is the solution to the system of equations.

For the first equation: \( 2x - y = 1 \)
We can rewrite this to find \( y \):
\( y = 2x - 1 \)
Let's find some points for this line:

For the second equation: \( x + 2y = 8 \)
We can rewrite this to find \( y \):
\( 2y = 8 - x \)
\( y = \frac {8 - x}{2} \)
Let's find some points for this line:

Now, we plot these points on a graph paper and draw the lines. The point where the two lines intersect is the solution.
The lines intersect at the point \( (2, 3) \).
The y-axis intersection for the first line (\( 2x - y = 1 \)) occurs when \( x = 0 \), which gives \( y = -1 \). So, the point is \( (0, -1) \).
The y-axis intersection for the second line (\( x + 2y = 8 \)) occurs when \( x = 0 \), which gives \( 2y = 8 \), so \( y = 4 \). So, the point is \( (0, 4) \).

Therefore, the required solution for the system of equations is \( x = 2 \) and \( y = 3 \). The lines intersect the y-axis at points \( (0, -1) \) and \( (0, 4) \) respectively.
In simple words: First, rewrite each equation to find \( y \) by itself. Then, pick some \( x \) values to find matching \( y \) values for both equations, making two tables. Plot these points on a graph and draw a line for each equation. The spot where the lines cross tells you the \( x \) and \( y \) values that solve both equations. Also, see where each line touches the up-and-down (y) axis.

🎯 Exam Tip: Always check your graphical solution by substituting the intersection point coordinates back into both original equations to ensure they are satisfied. For finding y-intercepts, simply set \( x = 0 \) in each equation.