PAIR OF LINEAR EQUATIONS IN TWO VARIABLES-Exercise 3.2
Maths - Exercise
Q.1 Solve the following pair of linear equations by the substitution method.
- \(x + y = 14,\)
\(x – y = 4\) - \(s – t = 3\)
\(\frac{s}{3} + \frac{t}{2}=6\) - \( 3x – y = 3\)
\( 9x – 3y = 9\) - \(0.2x + 0.3y = 1.3\)
\(0.4x + 0.5y = 2.3\) - \(\sqrt{2}x+\sqrt{3}y=0\)
\(\sqrt{3}x+\sqrt{8}y=0\) - \(\frac{3}{2}x-\frac{5}{3}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{16}\)
-
We are given the system of linear equations:
\[ x + y = 14 \tag{1} \] \[ x - y = 4 \tag{2} \]From equation (2), solve for \( y \) in terms of \( x \):
\[\begin{align} x - y = 4 \\\implies y = x - 4 \tag{3} \end{align}\]Substitute the expression for \( y \) from equation (3) into equation (1):
\[ x + (x - 4) = 14 \]Simplify and solve for \( x \):
\[ 2x - 4 = 14 \] \[ 2x = 18 \] \[ x = 9 \]Now, substitute \( x = 9 \) back into equation (3) to find \( y \):
\[ y = 9 - 4 = 5 \]Hence, the solution to the system is \( \boxed{x=9, \ y=5} \)
Verification: Substitute \( x=9 \), \( y=5 \) into original equations:
Equation (1): \( 9 + 5 = 14 \) ✓
Equation (2): \( 9 - 5 = 4 \) ✓ -
We are given the system of linear equations:
\[ s - t = 3 \tag{1} \] \[ \frac{s}{3} + \frac{t}{2} = 6 \tag{2} \]From equation (1), solve for \( s \) in terms of \( t \):
\[ s = t + 3 \tag{3} \]Substitute the expression for \( s \) from equation (3) into equation (2):
\[ \frac{t + 3}{3} + \frac{t}{2} = 6 \]To eliminate the fractions, multiply through by 6 (the least common multiple of 3 and 2):
\[ 2(t + 3) + 3t = 36 \]Simplify step by step:
\[ 2t + 6 + 3t = 36 \] \[ 5t + 6 = 36 \] \[ 5t = 30 \] \[ t = 6 \]Substitute \( t = 6 \) back into equation (3) to find \( s \):
\[ s = 6 + 3 = 9 \]Hence, the solution to the system is \( \boxed{s = 9,\ t = 6} \)
Verification: Equation (1): \( 9 - 6 = 3 \) ✓
Equation (2): \( \frac{9}{3} + \frac{6}{2} = 3 + 3 = 6 \) ✓ -
We are given the system of linear equations:
\[ 3x - y = 3 \tag{1} \] \[ 9x - 3y = 9 \tag{2} \]From equation (1), solve for \( y \) in terms of \( x \):
\[ y = 3x - 3 \tag{3} \]Substitute the expression for \( y \) from equation (3) into equation (2):
\[ 9x - 3(3x - 3) = 9 \]Simplify the left side step by step:
\[ 9x - 9x + 9 = 9 \] \[ 9 = 9 \]The equation \( 9 = 9 \) is always true for any value of \( x \). This indicates that equation (2) is simply three times equation (1), making the two equations identical.
Therefore, the lines are coincident, representing infinitely many solutions. The general solution is all points satisfying \( y = 3x - 3 \).
\[\scriptsize \boxed{\text{ Infinitely many solutions: }y = 3x - 3}\]Explanation: Equation (2) = 3 × Equation (1), so both represent the same straight line.
-
We are given the system of linear equations with decimal coefficients:
\[ 0.2x + 0.3y = 1.3 \tag{1} \] \[ 0.4x + 0.5y = 2.3 \tag{2} \]To eliminate decimals, multiply equation (1) by 10:
\[ 2x + 3y = 13 \]Solve for \( x \) in terms of \( y \):
\[ 2x = 13 - 3y \] \[ x = \frac{13 - 3y}{2} \tag{3} \]Multiply equation (2) by 10 to eliminate decimals:
\[ 4x + 5y = 23 \tag{4} \]Substitute the expression for \( x \) from equation (3) into equation (4):
\[ 4\left( \frac{13 - 3y}{2} \right) + 5y = 23 \]Simplify step by step:
\[ 2(13 - 3y) + 5y = 23 \] \[ 26 - 6y + 5y = 23 \] \[ 26 - y = 23 \] \[ -y = -3 \] \[ y = 3 \]Substitute \( y = 3 \) back into equation (3) to find \( x \):
\[\begin{aligned} x &= \frac{13 - 3(3)}{2} \\&= \frac{13 - 9}{2} \\&= \frac{4}{2} \\&= 2 \end{aligned}\]Hence, the solution to the system is \( \boxed{x = 2,\ y = 3} \)
Verification: Equation (1): \[\begin{aligned} 0.2(2) + 0.3(3) &= 0.4 + 0.9 \\&= 1.3 \end{aligned}\] ✓
Equation (2): \[\begin{aligned} 0.4(2) + 0.5(3) &= 0.8 + 1.5 \\&= 2.3 \end{aligned}\] ✓ -
We are given the pair of linear equations involving surds:
\[ \sqrt{2}\,x + \sqrt{3}\,y = 0 \tag{1} \] \[ \sqrt{3}\,x + \sqrt{8}\,y = 0 \tag{2} \]From equation (1), express \( x \) in terms of \( y \):
\[ \sqrt{2}\,x = -\sqrt{3}\,y \] \[\begin{align} x &= -\frac{\sqrt{3}}{\sqrt{2}}\,y \\&= -\sqrt{\frac{3}{2}}\,y \tag{3} \end{align}\]Substitute this value of \( x \) from equation (3) into equation (2):
\[ \sqrt{3}\left(-\sqrt{\frac{3}{2}}\,y\right) + \sqrt{8}\,y = 0 \]Simplify each term carefully:
\[ -\sqrt{3}\,\sqrt{\frac{3}{2}}\,y + \sqrt{8}\,y = 0 \] \[ -\sqrt{\frac{9}{2}}\,y + \sqrt{8}\,y = 0 \] \[ -\frac{3}{\sqrt{2}}\,y + 2\sqrt{2}\,y = 0 \]Take \( y \) common:
\[ y\left(-\frac{3}{\sqrt{2}} + 2\sqrt{2}\right) = 0 \]Simplify the expression in the bracket:
\[ 2\sqrt{2} = \frac{4}{\sqrt{2}} \] \[ -\frac{3}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{1}{\sqrt{2}} \neq 0 \]Thus the bracket is non-zero, so the only possibility is:
\[ y = 0 \]Substitute \( y = 0 \) into equation (1):
\[\begin{aligned} \sqrt{2}\,x + \sqrt{3}\cdot 0 &= 0 \\\implies \sqrt{2}\,x &= 0 \\\implies x &= 0 \end{aligned}\]Therefore, the pair of equations has the unique solution \( x = 0,\ y = 0 \)
\[ \boxed{x = 0,\ y = 0} \]Note: Both equations pass through the origin, and the substitution method confirms that \((0,0)\) is the only common solution.
-
We are given the pair of linear equations:
\(\displaystyle \frac{3}{2}x - \frac{5}{3}y = -2 \tag{1}\)
\(\displaystyle \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \tag{2}\)First, clear the denominators in equation (1) by multiplying both sides by 6 (the LCM of 2 and 3):
\[ 6\left(\frac{3}{2}x\right) - 6\left(\frac{5}{3}y\right) = 6(-2) \] \[ 9x - 10y = -12 \]Now express \( x \) in terms of \( y \):
\[ 9x = -12 + 10y \] \[ x = \frac{10y - 12}{9} \tag{3} \]Next, simplify equation (2). Combine the fractions on the left-hand side over a common denominator 6:
\[\begin{aligned} \frac{x}{3} + \frac{y}{2} &= \frac{2x}{6} + \frac{3y}{6} \\\\&= \frac{2x + 3y}{6} \\\\&= \frac{13}{6} \end{aligned}\] \[ 2x + 3y = 13 \tag{4} \]Substitute the value of \( x \) from equation (3) into equation (4):
\[ 2\left(\frac{10y - 12}{9}\right) + 3y = 13 \]Multiply through to simplify:
\[ \frac{20y - 24}{9} + 3y = 13 \]Multiply both sides by 9 to clear the denominator:
\[ 20y - 24 + 27y = 117 \] \[ 47y - 24 = 117 \] \[ 47y = 141 \] \[ y = \frac{141}{47} = 3 \]Now substitute \( y = 3 \) into equation (3) to find \( x \):
\[\begin{aligned} x &= \frac{10(3) - 12}{9} \\\\&= \frac{30 - 12}{9} \\\\&= \frac{18}{9} \\\\&= 2 \end{aligned}\]Therefore, the solution of the given pair of linear equations is \( \boxed{x = 2,\ y = 3} \)
Verification: In (1): \( \frac{3}{2}(2) - \frac{5}{3}(3) = 3 - 5 = -2 \) ✓
In (2): \( \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6} \) ✓
Q.2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution
First solve the pair of linear equations \(2x + 3y = 11\) and \(2x - 4y = -24\) by the substitution method, and then use the solution to find the value of \( m \) in the equation \( y = mx + 3 \).
From the first equation, express \(2x\) in terms of \( y \):
\[\begin{align} 2x + 3y &= 11 \\\implies 2x &= 11 - 3y \tag{1} \end{align}\]Substitute this value of \(2x\) into the second equation \(2x - 4y = -24\):
\[ (11 - 3y) - 4y = -24 \]Simplify step by step:
\[ 11 - 7y = -24 \] \[ -7y = -24 - 11 = -35 \] \[ y = \frac{-35}{-7} = 5 \]Now substitute \( y = 5 \) back into equation (1) to find \( x \):
\[ 2x = 11 - 3(5) = 11 - 15 = -4 \] \[ x = \frac{-4}{2} = -2 \]Thus, the solution of the pair of equations is \( x = -2 \) and \( y = 5 \).
Now use the relation \( y = mx + 3 \). Substitute \( x = -2 \) and \( y = 5 \) to find \( m \):
\[ 5 = m(-2) + 3 \] \[ -2m = 5 - 3 = 2 \] \[ m = -1 \]Therefore, \( x = -2 \), \( y = 5 \) and the required value of \( m \) is \( \boxed{-1} \)
Verification: Check in \( y = mx + 3 \): \( 5 = (-1)(-2) + 3 = 2 + 3 = 5 \) ✓
Q.3 Form the pair of linear equations for the following problems and find their solution by substitution method.
- The difference between two numbers is 26 and one number is three times the other. Find them.
- The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
- The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
- The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
- A fraction becomes \(\frac{9}{11}\) , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\) . Find the fraction.
- Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solutions
-
Solution
Let the smaller number be \( y \) and the larger number be \( x \). According to the problem, we know that one number is three times the other and their difference is 26.
This gives us the following equations:
\[ x = 3y \tag{1} \] \[ x - y = 26 \tag{2} \]Substitute the value of \( x \) from equation (1) into equation (2):
\[ 3y - y = 26 \]Simplify the left side:
\[ 2y = 26 \]Divide both sides by 2 to find \( y \):
\[ y = \frac{26}{2} = 13 \]Now substitute \( y = 13 \) back into equation (1) to find \( x \):
\[ x = 3 \times 13 = 39 \]Thus, the two numbers are 39 and 13.
\[ \boxed{39,\ 13} \]Verification: Difference = \( 39 - 13 = 26 \) ✓ and \( 39 = 3 \times 13 \) ✓
-
Solution
Let the smaller angle be \( y \) degrees and the larger angle be \( x \) degrees. Since the angles are supplementary, their sum must be 180 degrees.
This gives us the first equation:
\[ x + y = 180^\circ \tag{1} \]The problem also states that the larger angle exceeds the smaller by 18 degrees, so we have our second equation:
\[ x = y + 18^\circ \tag{2} \]Substitute the expression for \( x \) from equation (2) into equation (1):
\[ (y + 18^\circ) + y = 180^\circ \]Combine like terms and simplify:
\[ 2y + 18^\circ = 180^\circ \] \[ 2y = 180^\circ - 18^\circ \] \[ 2y = 162^\circ \] \[ y = \frac{162^\circ}{2} = 81^\circ \]Now substitute \( y = 81^\circ \) back into equation (2) to find \( x \):
\[ x = 81^\circ + 18^\circ = 99^\circ \]Therefore, the two supplementary angles are 99° and 81°.
\[ \boxed{99^\circ,\ 81^\circ} \]Verification: Sum = \( 99^\circ + 81^\circ = 180^\circ \) ✓ and difference = \( 99^\circ - 81^\circ = 18^\circ \) ✓
-
Solution
Let the cost of each bat be \( x \) rupees and the cost of each ball be \( y \) rupees. The coach bought 7 bats and 6 balls for ₹3800, which gives us our first equation.
This translates to:
\[ 7x + 6y = 3800 \tag{1} \]Later, she bought 3 bats and 5 balls for ₹1750, giving us the second equation:
\[ 3x + 5y = 1750 \tag{2} \]Solve equation (1) for \( x \):
\[ 7x = 3800 - 6y \] \[ x = \frac{3800 - 6y}{7} \]Substitute this expression for \( x \) into equation (2):
\[ 3\left( \frac{3800 - 6y}{7} \right) + 5y = 1750 \]Multiply through by 7 to eliminate the denominator:
\[ 3(3800 - 6y) + 35y = 1750 \times 7 \] \[ 11400 - 18y + 35y = 12250 \] \[ 17y = 850 \] \[ y = \frac{850}{17} = 50 \]Now substitute \( y = 50 \) back into equation (2) to find \( x \):
\[ 3x + 5(50) = 1750 \] \[ 3x + 250 = 1750 \] \[ 3x = 1500 \] \[ x = \frac{1500}{3} = 500 \]Therefore, the cost of each bat is ₹500 and each ball costs ₹50.
\[ \boxed{\text{Bat: ₹500, Ball: ₹50}} \]Verification: First purchase: \[\begin{aligned} 7(500) + 6(50) &= 3500 + 300 \\&= 3800 \text{ ✓}\end{aligned}\] Second purchase: \[\begin{aligned} 3(500) + 5(50) &= 1500 + 250 \\&= 1750 \text{ ✓}\end{aligned}\]
-
Solution
Let the fixed charge be \( x \) rupees and the charge per km be \( y \) rupees. For a 10 km journey, the total charge is ₹105, which gives us our first equation.
This translates to:
\[ x + 10y = 105 \tag{1} \]For a 15 km journey, the total charge is ₹155, giving us the second equation:
\[ x + 15y = 155 \tag{2} \]Solve equation (1) for \( x \):
\[ x = 105 - 10y \]Substitute this expression for \( x \) into equation (2):
\[ (105 - 10y) + 15y = 155 \]Simplify the equation step by step:
\[ 105 + 5y = 155 \] \[ 5y = 155 - 105 \] \[ 5y = 50 \] \[ y = \frac{50}{5} = 10 \]Now substitute \( y = 10 \) back into equation (1) to find \( x \):
\[ x + 10(10) = 105 \] \[ x + 100 = 105 \] \[ x = 5 \]Thus, the fixed charge is ₹5 and the charge per km is ₹10. For a 25 km journey, the total cost would be:
\[\begin{aligned} x + 25y &= 5 + 25(10) \\&= 5 + 250 \\&= 255 \end{aligned}\]Therefore, fixed charge = ₹5, charge per km = ₹10, and cost for 25 km = ₹255.
\[\scriptsize \boxed{\text{Fixed: ₹5, Per km: ₹10, 25 km: ₹255}}\]Verification: 10 km: \( 5 + 10(10) = 105 \) ✓
15 km: \( 5 + 15(10) = 155 \) ✓
25 km: \( 5 + 25(10) = 255 \) ✓ -
Solution
Let the fraction be \( \frac{x}{y} \) where \( x \) is the numerator and \( y \) is the denominator. When 2 is added to both, the fraction becomes \( \frac{9}{11} \).
This gives the equation:
\[ \frac{x + 2}{y + 2} = \frac{9}{11} \]Cross multiplying, we get:
\[ 11(x + 2) = 9(y + 2) \] \[ 11x + 22 = 9y + 18 \] \[ 11x = 9y - 4 \] \[ x = \frac{9y - 4}{11} \tag{1} \]Similarly, when 3 is added to both numerator and denominator, the fraction becomes \( \frac{5}{6} \), leading to:
\[ \frac{x + 3}{y + 3} = \frac{5}{6} \]Cross multiplying, we have:
\[ 6(x + 3) = 5(y + 3) \] \[ 6x + 18 = 5y + 15 \] \[ 6x - 5y = -3 \tag{2} \]Substituting \( x \) from equation (1) into equation (2):
\[ 6 \left( \frac{9y - 4}{11} \right) - 5y = -3 \] \[ \frac{54y - 24}{11} - 5y = -3 \]Multiply both sides by 11 to eliminate the denominator:
\[ 54y - 24 - 55y = -33 \] \[ -y - 24 = -33 \] \[ -y = -9 \] \[ y = 9 \]Substitute \( y = 9 \) into equation (1) to find \( x \):
\[\begin{aligned} x &= \frac{9 \times 9 - 4}{11} \\\\&= \frac{81 - 4}{11} \\\\&= \frac{77}{11} \\\\&= 7 \end{aligned}\]Therefore, the required fraction is \(\frac{7}{9}\).
\[ \boxed{\frac{7}{9}} \]Verification: Adding 2 to numerator and denominator: \(\frac{7+2}{9+2} = \frac{9}{11}\) ✓
Adding 3 to numerator and denominator: \(\frac{7+3}{9+3} = \frac{5}{6}\) ✓ -
Solution
Let Jacob's present age be \( x \) years and his son's present age be \( y \) years. Five years ago, Jacob's age was seven times his son's age.
This gives the first equation:
\[ x - 5 = 7(y - 5) \] \[ x - 5 = 7y - 35 \] \[ x = 7y - 30 \tag{1} \]Five years from now, Jacob's age will be three times his son's age, leading to the second equation:
\[ x + 5 = 3(y + 5) \] \[ x + 5 = 3y + 15 \tag{2} \]Substitute the expression for \( x \) from equation (1) into equation (2):
\[ (7y - 30) + 5 = 3y + 15 \]Simplify step by step:
\[ 7y - 25 = 3y + 15 \] \[ 7y - 3y = 15 + 25 \] \[ 4y = 40 \] \[ y = 10 \]Substitute \( y = 10 \) back into equation (1) to find \( x \):
\[\begin{aligned} x &= 7(10) - 30 \\&= 70 - 30 \\&= 40 \end{aligned}\]Therefore, Jacob's present age is 40 years and his son's present age is 10 years.
\[\scriptsize \boxed{\text{Jacob: 40 years, Son: 10 years}}\]Verification: 5 years ago: Jacob = 35, Son = 5, and \( 35 = 7 \times 5 \) ✓
5 years hence: Jacob = 45, Son = 15, and \( 45 = 3 \times 15 \) ✓
