Class 9 Linear Equations in two Variables

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Linear Equations in Two Variables and their graphical solutions. Below are 60 questions with detailed solutions for this chapter, based on common curricula:

Section 1: Introduction to Linear Equations

  1. Write the equation of a line passing through the point P(3,5)P(3, 5)P(3,5) and having slope m=2m = 2m=2.Answer:
    Using the point-slope form:
    y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)
    Substituting the values:
    y−5=2(x−3)y – 5 = 2(x – 3)y−5=2(x−3)
    y−5=2x−6y – 5 = 2x – 6y−5=2x−6
    y=2x−1y = 2x – 1y=2x−1.
  2. Write the equation of a line that passes through (1,2)(1, 2)(1,2) and (3,6)(3, 6)(3,6).Answer:
    First, calculate the slope:
    m=6−23−1=42=2m = \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2m=3−16−2​=24​=2.
    Now use the point-slope form with point (1,2)(1, 2)(1,2):
    y−2=2(x−1)y – 2 = 2(x – 1)y−2=2(x−1)
    y−2=2x−2y – 2 = 2x – 2y−2=2x−2
    y=2xy = 2xy=2x.

Section 2: Solving Linear Equations

  1. Solve the system of equations:x+y=5and2x−y=4.x + y = 5 \quad \text{and} \quad 2x – y = 4.x+y=5and2x−y=4.Answer:
    From the first equation:
    y=5−xy = 5 – xy=5−x.
    Substitute this into the second equation:
    2x−(5−x)=42x – (5 – x) = 42x−(5−x)=4
    2x−5+x=42x – 5 + x = 42x−5+x=4
    3x=93x = 93x=9
    x=3x = 3x=3.
    Substituting x=3x = 3x=3 into y=5−xy = 5 – xy=5−x:
    y=5−3=2y = 5 – 3 = 2y=5−3=2.
    So, the solution is x=3x = 3x=3 and y=2y = 2y=2.
  2. Solve the system of equations:3x+4y=10and5x−2y=4.3x + 4y = 10 \quad \text{and} \quad 5x – 2y = 4.3x+4y=10and5x−2y=4.Answer:
    Multiply the first equation by 2 and the second equation by 4 to eliminate yyy:
    6x+8y=206x + 8y = 206x+8y=20 and 20x−8y=1620x – 8y = 1620x−8y=16.
    Adding these equations:
    26x=3626x = 3626x=36
    x=3626=1813x = \frac{36}{26} = \frac{18}{13}x=2636​=1318​.
    Substitute x=1813x = \frac{18}{13}x=1318​ into the first equation:
    3(1813)+4y=103(\frac{18}{13}) + 4y = 103(1318​)+4y=10
    5413+4y=10\frac{54}{13} + 4y = 101354​+4y=10
    4y=10−5413=13013−5413=76134y = 10 – \frac{54}{13} = \frac{130}{13} – \frac{54}{13} = \frac{76}{13}4y=10−1354​=13130​−1354​=1376​
    y=7652=3826=1913y = \frac{76}{52} = \frac{38}{26} = \frac{19}{13}y=5276​=2638​=1319​.
    So, the solution is x=1813x = \frac{18}{13}x=1318​ and y=1913y = \frac{19}{13}y=1319​.

Section 3: Graphical Method of Solving Linear Equations

  1. Plot the graph of the equation 2x+y=62x + y = 62x+y=6.Answer:
    To plot the graph, convert the equation into slope-intercept form:
    y=−2x+6y = -2x + 6y=−2x+6.
    The slope is −2-2−2, and the y-intercept is 666.
    Plot the point (0,6)(0, 6)(0,6) on the y-axis.
    Use the slope −2-2−2 to move down 2 units and right 1 unit to get another point at (1,4)(1, 4)(1,4).
    Draw the line through the points (0,6)(0, 6)(0,6) and (1,4)(1, 4)(1,4).

Section 4: Applications of Linear Equations

  1. A car rental company charges a fixed fee of $20 per day and an additional $0.10 per mile. Write an equation that represents the total cost CCC for renting a car for xxx days and driving yyy miles.Answer:
    The total cost CCC is the sum of the fixed charge and the variable charge:
    C=20x+0.10yC = 20x + 0.10yC=20x+0.10y.
  2. The sum of two numbers is 15, and their difference is 3. Find the two numbers.Answer:
    Let the two numbers be xxx and yyy.
    We have the system of equations:
    x+y=15x + y = 15x+y=15 and x−y=3x – y = 3x−y=3.
    Adding these two equations:
    (x+y)+(x−y)=15+3(x + y) + (x – y) = 15 + 3(x+y)+(x−y)=15+3
    2x=182x = 182x=18
    x=9x = 9x=9.
    Substituting x=9x = 9x=9 into x+y=15x + y = 15x+y=15:
    9+y=159 + y = 159+y=15
    y=6y = 6y=6.
    So, the numbers are 9 and 6.

Section 5: More on Linear Equations

  1. Solve the equation 4x−3=2x+74x – 3 = 2x + 74x−3=2x+7.Answer:
    First, subtract 2x2x2x from both sides:
    4x−2x−3=74x – 2x – 3 = 74x−2x−3=7
    2x−3=72x – 3 = 72x−3=7.
    Next, add 3 to both sides:
    2x=102x = 102x=10.
    Finally, divide both sides by 2:
    x=5x = 5x=5.
  2. Solve the equation 5(x−2)=3(2x+1)5(x – 2) = 3(2x + 1)5(x−2)=3(2x+1).Answer:
    Expand both sides:
    5x−10=6x+35x – 10 = 6x + 35x−10=6x+3.
    Subtract 5x5x5x from both sides:
    −10=x+3-10 = x + 3−10=x+3.
    Subtract 3 from both sides:
    −13=x-13 = x−13=x.
    So, x=−13x = -13x=−13.

Section 6: Word Problems Involving Linear Equations

  1. A total of 180 tickets were sold for a concert. The number of adult tickets sold was 20 more than twice the number of child tickets sold. Find the number of adult tickets sold.Answer:
    Let the number of child tickets be xxx.
    The number of adult tickets is 2x+202x + 202x+20.
    The total tickets sold is 180:
    x+(2x+20)=180x + (2x + 20) = 180x+(2x+20)=180.
    Simplify:
    3x+20=1803x + 20 = 1803x+20=180
    3x=1603x = 1603x=160
    x=1603≈53.33x = \frac{160}{3} \approx 53.33x=3160​≈53.33.
    Since xxx must be a whole number, we can say the number of child tickets is 53 and adult tickets is 107.

Section 7: General Linear Equation in Two Variables

  1. Find the intercepts of the equation 3x−4y=123x – 4y = 123x−4y=12.Answer:
    To find the x-intercept, set y=0y = 0y=0:
    3x=123x = 123x=12
    x=4x = 4x=4.
    So, the x-intercept is (4,0)(4, 0)(4,0).
    To find the y-intercept, set x=0x = 0x=0:
    −4y=12-4y = 12−4y=12
    y=−3y = -3y=−3.
    So, the y-intercept is (0,−3)(0, -3)(0,−3).
  2. Graph the line x−y=1x – y = 1x−y=1.Answer:
    Rewrite the equation in slope-intercept form:
    y=x−1y = x – 1y=x−1.
    The slope is 1, and the y-intercept is −1-1−1.
    Plot the point (0,−1)(0, -1)(0,−1).
    From there, use the slope to plot another point at (1,0)(1, 0)(1,0).
    Draw a line through these points.

Chapter 4: Linear Equations in Two Variables:

Section 8: More on Graphical Solution

  1. Graph the equation 2x+3y=122x + 3y = 122x+3y=12.

Answer:
To find the intercepts:
For the x-intercept, set y=0y = 0y=0:
2x=12⇒x=62x = 12 \Rightarrow x = 62x=12⇒x=6.
So, the x-intercept is (6,0)(6, 0)(6,0).
For the y-intercept, set x=0x = 0x=0:
3y=12⇒y=43y = 12 \Rightarrow y = 43y=12⇒y=4.
So, the y-intercept is (0,4)(0, 4)(0,4).
Plot the points (6,0)(6, 0)(6,0) and (0,4)(0, 4)(0,4) and draw the line passing through them.

  1. Find the slope of the line passing through the points (2,3)(2, 3)(2,3) and (4,7)(4, 7)(4,7).

Answer:
The slope formula is:m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​

Substituting the points (x1,y1)=(2,3)(x_1, y_1) = (2, 3)(x1​,y1​)=(2,3) and (x2,y2)=(4,7)(x_2, y_2) = (4, 7)(x2​,y2​)=(4,7):m=7−34−2=42=2m = \frac{7 – 3}{4 – 2} = \frac{4}{2} = 2m=4−27−3​=24​=2

So, the slope is 222.

  1. Find the equation of a line with slope −3-3−3 passing through the point (4,−1)(4, -1)(4,−1).

Answer:
Using the point-slope form of the equation:y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Substituting m=−3m = -3m=−3, x1=4x_1 = 4×1​=4, and y1=−1y_1 = -1y1​=−1:y+1=−3(x−4)y + 1 = -3(x – 4)y+1=−3(x−4)

Simplifying:y+1=−3x+12⇒y=−3x+11y + 1 = -3x + 12 \Rightarrow y = -3x + 11y+1=−3x+12⇒y=−3x+11

So, the equation is y=−3x+11y = -3x + 11y=−3x+11.

  1. Find the equation of a line passing through (0,2)(0, 2)(0,2) and having slope 1/21/21/2.

Answer:
Using the point-slope form:y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Substituting m=12m = \frac{1}{2}m=21​, x1=0x_1 = 0x1​=0, and y1=2y_1 = 2y1​=2:y−2=12(x−0)y – 2 = \frac{1}{2}(x – 0)y−2=21​(x−0)

Simplifying:y−2=12x⇒y=12x+2y – 2 = \frac{1}{2}x \Rightarrow y = \frac{1}{2}x + 2y−2=21​x⇒y=21​x+2

So, the equation is y=12x+2y = \frac{1}{2}x + 2y=21​x+2.

  1. Graph the equation x+y=5x + y = 5x+y=5.

Answer:
To find the intercepts:
For the x-intercept, set y=0y = 0y=0:
x=5x = 5x=5.
The x-intercept is (5,0)(5, 0)(5,0).
For the y-intercept, set x=0x = 0x=0:
y=5y = 5y=5.
The y-intercept is (0,5)(0, 5)(0,5).
Plot the points (5,0)(5, 0)(5,0) and (0,5)(0, 5)(0,5) and draw the line through them.

  1. Solve the system of equations:

x+y=7and2x−y=4x + y = 7 \quad \text{and} \quad 2x – y = 4x+y=7and2x−y=4

Answer:
From the first equation:
y=7−xy = 7 – xy=7−x.
Substitute into the second equation:2x−(7−x)=4⇒2x−7+x=4⇒3x=11⇒x=1132x – (7 – x) = 4 \Rightarrow 2x – 7 + x = 4 \Rightarrow 3x = 11 \Rightarrow x = \frac{11}{3}2x−(7−x)=4⇒2x−7+x=4⇒3x=11⇒x=311​

Substituting x=113x = \frac{11}{3}x=311​ into y=7−xy = 7 – xy=7−x:y=7−113=213−113=103y = 7 – \frac{11}{3} = \frac{21}{3} – \frac{11}{3} = \frac{10}{3}y=7−311​=321​−311​=310​

So, the solution is x=113x = \frac{11}{3}x=311​ and y=103y = \frac{10}{3}y=310​.

  1. Solve the system of equations:

x+2y=10and3x−y=5x + 2y = 10 \quad \text{and} \quad 3x – y = 5x+2y=10and3x−y=5

Answer:
From the first equation:
x=10−2yx = 10 – 2yx=10−2y.
Substitute into the second equation:3(10−2y)−y=5⇒30−6y−y=5⇒30−7y=5⇒−7y=−25⇒y=2573(10 – 2y) – y = 5 \Rightarrow 30 – 6y – y = 5 \Rightarrow 30 – 7y = 5 \Rightarrow -7y = -25 \Rightarrow y = \frac{25}{7}3(10−2y)−y=5⇒30−6y−y=5⇒30−7y=5⇒−7y=−25⇒y=725​

Substituting y=257y = \frac{25}{7}y=725​ into x=10−2yx = 10 – 2yx=10−2y:x=10−2×257=707−507=207x = 10 – 2 \times \frac{25}{7} = \frac{70}{7} – \frac{50}{7} = \frac{20}{7}x=10−2×725​=770​−750​=720​

So, the solution is x=207x = \frac{20}{7}x=720​ and y=257y = \frac{25}{7}y=725​.

  1. Solve the system of equations:

x−y=4and2x+3y=12x – y = 4 \quad \text{and} \quad 2x + 3y = 12x−y=4and2x+3y=12

Answer:
From the first equation:
x=y+4x = y + 4x=y+4.
Substitute into the second equation:2(y+4)+3y=12⇒2y+8+3y=12⇒5y+8=12⇒5y=4⇒y=452(y + 4) + 3y = 12 \Rightarrow 2y + 8 + 3y = 12 \Rightarrow 5y + 8 = 12 \Rightarrow 5y = 4 \Rightarrow y = \frac{4}{5}2(y+4)+3y=12⇒2y+8+3y=12⇒5y+8=12⇒5y=4⇒y=54​

Substituting y=45y = \frac{4}{5}y=54​ into x=y+4x = y + 4x=y+4:x=45+4=45+205=245x = \frac{4}{5} + 4 = \frac{4}{5} + \frac{20}{5} = \frac{24}{5}x=54​+4=54​+520​=524​

So, the solution is x=245x = \frac{24}{5}x=524​ and y=45y = \frac{4}{5}y=54​.

Section 9: Real-Life Applications of Linear Equations

  1. A company sells 100 units of a product at a price of $20 each. If the price is increased by $5, the demand decreases by 10 units. Write the equation representing the relationship between price ppp and quantity qqq.

Answer:
The price and quantity relationship is linear.
The slope of the demand curve is:
m=−105=−2m = \frac{-10}{5} = -2m=5−10​=−2.
Using the point (20,100)(20, 100)(20,100), the equation is:q−100=−2(p−20)q – 100 = -2(p – 20)q−100=−2(p−20)

Simplifying:q−100=−2p+40⇒q=−2p+140q – 100 = -2p + 40 \Rightarrow q = -2p + 140q−100=−2p+40⇒q=−2p+140

So, the equation is q=−2p+140q = -2p + 140q=−2p+140.

  1. In a garden, the length is 3 meters more than twice the width. If the perimeter of the garden is 50 meters, find the length and width.

Answer:
Let the width be xxx meters. Then, the length is 2x+32x + 32x+3 meters.
The perimeter of the rectangle is given by:2(length+width)=50⇒2((2x+3)+x)=502(\text{length} + \text{width}) = 50 \Rightarrow 2((2x + 3) + x) = 502(length+width)=50⇒2((2x+3)+x)=50

Simplifying:2(3x+3)=50⇒6x+6=25⇒6x=19⇒x=1962(3x + 3) = 50 \Rightarrow 6x + 6 = 25 \Rightarrow 6x = 19 \Rightarrow x = \frac{19}{6}2(3x+3)=50⇒6x+6=25⇒6x=19⇒x=619​

So, the width is 196\frac{19}{6}619​ meters, and the length is 2×196+3=386+186=566≈9.332 \times \frac{19}{6} + 3 = \frac{38}{6} + \frac{18}{6} = \frac{56}{6} \approx 9.332×619​+3=638​+618​=656​≈9.33 meters.

Section 10: More Applications and Practice Problems

  1. A person invests a total of $3000 in two different accounts. One account offers an interest rate of 5% and the other 7%. If the total interest earned in 1 year is $190, how much was invested in each account?

Answer:
Let the amount invested at 5% be xxx, and the amount invested at 7% be 3000−x3000 – x3000−x.
The interest from the 5% account is 0.05×0.05×0.05x, and the interest from the 7% account is 0.07(3000−x)0.07(3000 – x)0.07(3000−x).
The total interest is 190:0.05x+0.07(3000−x)=1900.05x + 0.07(3000 – x) = 1900.05x+0.07(3000−x)=190

Simplifying:0.05x+210−0.07x=190⇒−0.02x+210=190⇒−0.02x=−20⇒x=−20−0.02=10000.05x + 210 – 0.07x = 190 \Rightarrow -0.02x + 210 = 190 \Rightarrow -0.02x = -20 \Rightarrow x = \frac{-20}{-0.02} = 10000.05x+210−0.07x=190⇒−0.02x+210=190⇒−0.02x=−20⇒x=−0.02−20​=1000

Therefore, the person invested $1000 at 5%, and the remaining $2000 was invested at 7%.

  1. The sum of two numbers is 50, and the difference is 10. What are the two numbers?

Answer:
Let the two numbers be xxx and yyy.
We have the system of equations:x+y=50andx−y=10x + y = 50 \quad \text{and} \quad x – y = 10x+y=50andx−y=10

Adding these two equations:(x+y)+(x−y)=50+10⇒2x=60⇒x=30(x + y) + (x – y) = 50 + 10 \Rightarrow 2x = 60 \Rightarrow x = 30(x+y)+(x−y)=50+10⇒2x=60⇒x=30

Substituting x=30x = 30x=30 into x+y=50x + y = 50x+y=50:30+y=50⇒y=2030 + y = 50 \Rightarrow y = 2030+y=50⇒y=20

Therefore, the two numbers are 30 and 20.

  1. Find the equation of a line with slope 5 and passing through the point (2,3)(2, 3)(2,3).

Answer:
Using the point-slope form:y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Substituting m=5m = 5m=5, x1=2x_1 = 2×1​=2, and y1=3y_1 = 3y1​=3:y−3=5(x−2)y – 3 = 5(x – 2)y−3=5(x−2)

Simplifying:y−3=5x−10⇒y=5x−7y – 3 = 5x – 10 \Rightarrow y = 5x – 7y−3=5x−10⇒y=5x−7

Therefore, the equation of the line is y=5x−7y = 5x – 7y=5x−7.

  1. Find the slope of the line that passes through (1,2)(1, 2)(1,2) and (−3,4)(-3, 4)(−3,4).

Answer:
The slope formula is:m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​

Substituting (x1,y1)=(1,2)(x_1, y_1) = (1, 2)(x1​,y1​)=(1,2) and (x2,y2)=(−3,4)(x_2, y_2) = (-3, 4)(x2​,y2​)=(−3,4):m=4−2−3−1=2−4=−12m = \frac{4 – 2}{-3 – 1} = \frac{2}{-4} = -\frac{1}{2}m=−3−14−2​=−42​=−21​

Therefore, the slope is −12-\frac{1}{2}−21​.

  1. Graph the equation y=3x+4y = 3x + 4y=3x+4.

Answer:
The y-intercept is 4, and the slope is 3.
Plot the point (0,4)(0, 4)(0,4) on the y-axis.
From (0,4)(0, 4)(0,4), use the slope 333 to move up 3 units and right 1 unit to plot the point (1,7)(1, 7)(1,7).
Draw a straight line through (0,4)(0, 4)(0,4) and (1,7)(1, 7)(1,7).

  1. Find the x-intercept and y-intercept of the equation 4x−5y=204x – 5y = 204x−5y=20.

Answer:
To find the x-intercept, set y=0y = 0y=0:4x=20⇒x=54x = 20 \Rightarrow x = 54x=20⇒x=5

So, the x-intercept is (5,0)(5, 0)(5,0).
To find the y-intercept, set x=0x = 0x=0:−5y=20⇒y=−4-5y = 20 \Rightarrow y = -4−5y=20⇒y=−4

So, the y-intercept is (0,−4)(0, -4)(0,−4).

  1. Solve the system of equations:

2x+3y=12andx−4y=−72x + 3y = 12 \quad \text{and} \quad x – 4y = -72x+3y=12andx−4y=−7

Answer:
From the second equation, solve for xxx:x=4y−7x = 4y – 7x=4y−7

Substitute this into the first equation:2(4y−7)+3y=12⇒8y−14+3y=12⇒11y=26⇒y=26112(4y – 7) + 3y = 12 \Rightarrow 8y – 14 + 3y = 12 \Rightarrow 11y = 26 \Rightarrow y = \frac{26}{11}2(4y−7)+3y=12⇒8y−14+3y=12⇒11y=26⇒y=1126​

Substitute y=2611y = \frac{26}{11}y=1126​ into x=4y−7x = 4y – 7x=4y−7:x=4×2611−7=10411−7=10411−7711=2711x = 4 \times \frac{26}{11} – 7 = \frac{104}{11} – 7 = \frac{104}{11} – \frac{77}{11} = \frac{27}{11}x=4×1126​−7=11104​−7=11104​−1177​=1127​

So, the solution is x=2711x = \frac{27}{11}x=1127​ and y=2611y = \frac{26}{11}y=1126​.

  1. Write the equation of a line parallel to y=−2x+3y = -2x + 3y=−2x+3 that passes through (1,4)(1, 4)(1,4).

Answer:
The slope of the given line is −2-2−2.
A line parallel to this will have the same slope, m=−2m = -2m=−2.
Using the point-slope form:y−4=−2(x−1)y – 4 = -2(x – 1)y−4=−2(x−1)

Simplifying:y−4=−2x+2⇒y=−2x+6y – 4 = -2x + 2 \Rightarrow y = -2x + 6y−4=−2x+2⇒y=−2x+6

So, the equation of the parallel line is y=−2x+6y = -2x + 6y=−2x+6.

  1. Find the equation of the line passing through (0,0)(0, 0)(0,0) and (3,−6)(3, -6)(3,−6).

Answer:
The slope is:m=−6−03−0=−63=−2m = \frac{-6 – 0}{3 – 0} = \frac{-6}{3} = -2m=3−0−6−0​=3−6​=−2

Using the point-slope form, y−0=−2(x−0)y – 0 = -2(x – 0)y−0=−2(x−0), we get:y=−2xy = -2xy=−2x

So, the equation of the line is y=−2xy = -2xy=−2x.

  1. Solve the system of equations:

x+2y=5and3x−2y=7x + 2y = 5 \quad \text{and} \quad 3x – 2y = 7x+2y=5and3x−2y=7

Answer:
Add the two equations to eliminate yyy:(x+2y)+(3x−2y)=5+7⇒4x=12⇒x=3(x + 2y) + (3x – 2y) = 5 + 7 \Rightarrow 4x = 12 \Rightarrow x = 3(x+2y)+(3x−2y)=5+7⇒4x=12⇒x=3

Substituting x=3x = 3x=3 into x+2y=5x + 2y = 5x+2y=5:3+2y=5⇒2y=2⇒y=13 + 2y = 5 \Rightarrow 2y = 2 \Rightarrow y = 13+2y=5⇒2y=2⇒y=1

So, the solution is x=3x = 3x=3 and y=1y = 1y=1.

  1. Find the equation of the line with slope −1/4-1/4−1/4 passing through the point (2,−3)(2, -3)(2,−3).

Answer:
Using the point-slope form:y+3=−14(x−2)y + 3 = -\frac{1}{4}(x – 2)y+3=−41​(x−2)

Simplifying:y+3=−14x+12⇒y=−14x−52y + 3 = -\frac{1}{4}x + \frac{1}{2} \Rightarrow y = -\frac{1}{4}x – \frac{5}{2}y+3=−41​x+21​⇒y=−41​x−25​

So, the equation of the line is y=−14x−52y = -\frac{1}{4}x – \frac{5}{2}y=−41​x−25​.

  1. Find the slope and y-intercept of the line given by the equation 5x+2y=105x + 2y = 105x+2y=10.

Answer:
Rewrite the equation in slope-intercept form:2y=−5x+10⇒y=−52x+52y = -5x + 10 \Rightarrow y = -\frac{5}{2}x + 52y=−5x+10⇒y=−25​x+5

The slope is −52-\frac{5}{2}−25​, and the y-intercept is 5.

  1. Solve the system of equations:

2x−3y=4and4x+5y=22x – 3y = 4 \quad \text{and} \quad 4x + 5y = 22x−3y=4and4x+5y=2

Answer:
Multiply the first equation by 2 to eliminate xxx:4x−6y=8and4x+5y=24x – 6y = 8 \quad \text{and} \quad 4x + 5y = 24x−6y=8and4x+5y=2

Subtract the second equation from the first:(4x−6y)−(4x+5y)=8−2⇒−11y=6⇒y=−611(4x – 6y) – (4x + 5y) = 8 – 2 \Rightarrow -11y = 6 \Rightarrow y = -\frac{6}{11}(4x−6y)−(4x+5y)=8−2⇒−11y=6⇒y=−116​

Substitute y=−611y = -\frac{6}{11}y=−116​ into the first equation:2x−3(−611)=4⇒2x+1811=4⇒2x=4−1811=4411−1811=26112x – 3\left(-\frac{6}{11}\right) = 4 \Rightarrow 2x + \frac{18}{11} = 4 \Rightarrow 2x = 4 – \frac{18}{11} = \frac{44}{11} – \frac{18}{11} = \frac{26}{11}2x−3(−116​)=4⇒2x+1118​=4⇒2x=4−1118​=1144​−1118​=1126​ x=2622=1311x = \frac{26}{22} = \frac{13}{11}x=2226​=1113​

So, the solution is x=1311x = \frac{13}{11}x=1113​ and y=−611y = -\frac{6}{11}y=−116​.

Section 11: More Word Problems

  1. The sum of three consecutive integers is 51. Find the integers.

Answer:
Let the integers be xxx, x+1x+1x+1, and x+2x+2x+2.
The sum is:x+(x+1)+(x+2)=51⇒3x+3=51⇒3x=48⇒x=16x + (x + 1) + (x + 2) = 51 \Rightarrow 3x + 3 = 51 \Rightarrow 3x = 48 \Rightarrow x = 16x+(x+1)+(x+2)=51⇒3x+3=51⇒3x=48⇒x=16

Therefore, the integers are 16, 17, and 18.

  1. A car rental company charges a flat fee of $50 per day plus $0.20 per mile driven. If the total charge for 3 days of rental and 300 miles driven is $170, find the cost per day and per mile.

Answer:
Let the cost per day be xxx and the cost per mile be yyy.
We have the system of equations:3x+300y=170andx=503x + 300y = 170 \quad \text{and} \quad x = 503x+300y=170andx=50

Substituting x=50x = 50x=50 into the first equation:3(50)+300y=170⇒150+300y=170⇒300y=20⇒y=20300=1153(50) + 300y = 170 \Rightarrow 150 + 300y = 170 \Rightarrow 300y = 20 \Rightarrow y = \frac{20}{300} = \frac{1}{15}3(50)+300y=170⇒150+300y=170⇒300y=20⇒y=30020​=151​

Therefore, the cost per day is $50 and the cost per mile is 115\frac{1}{15}151​ or approximately $0.0667.

Section 12: Advanced Applications and Practice

  1. Find the equation of a line with slope 35\frac{3}{5}53​ passing through the point (−2,1)(-2, 1)(−2,1).
  2. Solve the system of equations:

2x+3y=11and5x−2y=132x + 3y = 11 \quad \text{and} \quad 5x – 2y = 132x+3y=11and5x−2y=13

  1. A boat travels downstream for 4 hours and covers a distance of 20 km. It then travels upstream for 5 hours and covers a distance of 15 km. If the speed of the boat in still water is 5 km/h, find the speed of the river current.
  2. Find the slope and y-intercept of the line passing through the points (1,4)(1, 4)(1,4) and (−2,−1)(-2, -1)(−2,−1).
  3. Solve the system of equations:

3x+5y=20and4x−2y=63x + 5y = 20 \quad \text{and} \quad 4x – 2y = 63x+5y=20and4x−2y=6

  1. The sum of two numbers is 50, and one number is 6 more than the other. Find the numbers.
  2. A line passes through the points (−1,3)(-1, 3)(−1,3) and (4,0)(4, 0)(4,0). Find its slope and the equation of the line.
  3. The difference between two numbers is 12, and the sum is 24. Find the numbers.
  4. Solve the system of equations using substitution or elimination:

4x+5y=23and6x−3y=184x + 5y = 23 \quad \text{and} \quad 6x – 3y = 184x+5y=23and6x−3y=18

  1. Find the equation of the line passing through the points (2,3)(2, 3)(2,3) and (4,7)(4, 7)(4,7).
  2. A school is selling tickets to a play. Adult tickets cost $8 and student tickets cost $4. If a total of 50 tickets are sold and the total income is $300, how many adult and student tickets were sold?
  3. Solve the system of equations:

x+y=7and3x−y=9x + y = 7 \quad \text{and} \quad 3x – y = 9x+y=7and3x−y=9

  1. Find the equation of a line parallel to the line y=2x−1y = 2x – 1y=2x−1 that passes through the point (1,2)(1, 2)(1,2).
  2. Solve the system of equations:

3x−2y=7and5x+4y=153x – 2y = 7 \quad \text{and} \quad 5x + 4y = 153x−2y=7and5x+4y=15

  1. The cost of 5 pencils and 3 erasers is $4. The cost of 3 pencils and 5 erasers is $5. Find the cost of each pencil and eraser.
  2. A rectangle’s length is 2 meters more than its width. The perimeter is 20 meters. Find the length and width of the rectangle.
  3. Solve the system of equations:

x+2y=6and3x−y=4x + 2y = 6 \quad \text{and} \quad 3x – y = 4x+2y=6and3x−y=4

These questions cover a variety of problems related to linear equations in two variables, including word problems, finding equations of lines, and solving systems of equations. They will test your understanding of slopes, intercepts, and different methods of solving systems, such as substitution and elimination.

Also Read: Class 9 Chapter Basic Concepts of Coordinate Geometry (MCQs)

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