SIMULTANEOUS (Linear) EQUATIONS [Including Problems] - Questions & Answers
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) Solution of equations x - 2y = 7 and 2x - y = 8 is :
(i) x = 3, y = 2
(ii) x = -3, y = 2
(iii) x = 3, y = -2
(iv) x = -3, y = -2
Answer: (iii)
Step 1: The given equations are x - 2y = 7 ...(1) and 2x - y = 8 ...(2)
Step 2: From equation (1), express x in terms of y: x = 2y + 7
Step 3: Substitute this value of x in equation (2): 2(2y + 7) - y = 8
Step 4: Expand the equation: 4y + 14 - y = 8
Step 5: Simplify to find y: 3y = 8 - 14
Step 6: 3y = -6
Step 7: y = -2
Step 8: Substitute y = -2 in x = 2y + 7
Step 9: x = 2(-2) + 7 = -4 + 7 = 3
Step 10: Therefore, the solution is x = 3, y = -2. This matches option (iii).
(b) Solution of equations x + y = 2.1 and x - y = 0.3 is :
(i) x = 1.2, y = 0.9
(ii) x = 1.2, y = -0.9
(iii) x = -1.2, y = 0.9
(iv) x = -1.2, y = -0.9
Answer: (i)
Step 1: The given equations are x + y = 2.1 ...(1) and x - y = 0.3 ...(2)
Step 2: Add equation (1) and equation (2) together to eliminate y.
Step 3: (x + x) + (y - y) = 2.1 + 0.3
Step 4: 2x = 2.4
Step 5: x = 2.4 / 2 = 1.2
Step 6: Substitute x = 1.2 in equation (1): 1.2 + y = 2.1
Step 7: y = 2.1 - 1.2 = 0.9
Step 8: The solution is x = 1.2, y = 0.9. This matches option (i).
(c) Solution of equations 1/x + 1/y = 21 and 1/x - 1/y + 9 = 0 is :
(i) x = 1/6, y = 1/15
(ii) x = -1/6, y = 1/15
(iii) x = -1/6, y = -1/15
(iv) x = 1/6, y = -1/15
Answer: (i)
Step 1: Let 1/x = u and 1/y = v to simplify the equations.
Step 2: The equations become u + v = 21 ...(1) and u - v + 9 = 0 => u - v = -9 ...(2)
Step 3: Add equation (1) and equation (2): (u + v) + (u - v) = 21 + (-9)
Step 4: 2u = 12
Step 5: u = 12 / 2 = 6
Step 6: Since 1/x = u, we have 1/x = 6, which means x = 1/6.
Step 7: Substitute u = 6 into equation (1): 6 + v = 21
Step 8: v = 21 - 6 = 15
Step 9: Since 1/y = v, we have 1/y = 15, which means y = 1/15.
Step 10: The solution is x = 1/6, y = 1/15. This matches option (i).
(d) Solution of equations x/2 - y/3 = 0 and x/2 + y/3 = 6 is :
(i) x = 6, y = -9
(ii) x = 6, y = 9
(iii) x = -6, y = -9
(iv) x = -6, y = 9
Answer: (ii)
Step 1: The given equations are x/2 - y/3 = 0 ...(1) and x/2 + y/3 = 6 ...(2)
Step 2: Add equation (1) and equation (2) to eliminate the y terms.
Step 3: (x/2 + x/2) + (-y/3 + y/3) = 0 + 6
Step 4: x = 6
Step 5: Substitute x = 6 into equation (2): 6/2 + y/3 = 6
Step 6: 3 + y/3 = 6
Step 7: y/3 = 6 - 3 = 3
Step 8: y = 3 * 3 = 9
Step 9: The solution is x = 6, y = 9. This matches option (ii).
(e) Solution of equations (x+8)/5 - (2y-4)/3 = 0 and y - x = 3 is :
(i) x = 2, y = 5
(ii) x = -2, y = 5
(iii) x = 2, y = -5
(iv) x = -2, y = -5
Answer: (i)
Step 1: From the second equation y - x = 3, express y in terms of x: y = x + 3
Step 2: Substitute y = x + 3 into the first equation: (x+8)/5 - (2(x+3)-4)/3 = 0
Step 3: Simplify the numerator of the second fraction: 2x + 6 - 4 = 2x + 2
Step 4: The equation becomes (x+8)/5 - (2x+2)/3 = 0
Step 5: Multiply the entire equation by the LCM of 5 and 3, which is 15.
Step 6: 3(x + 8) - 5(2x + 2) = 0
Step 7: Expand the brackets: 3x + 24 - 10x - 10 = 0
Step 8: Group like terms: -7x + 14 = 0
Step 9: 7x = 14 => x = 2
Step 10: Substitute x = 2 into y = x + 3: y = 2 + 3 = 5
Step 11: The solution is x = 2, y = 5. This matches option (i).
Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:
2. 2x + 3y = 8
2x = 2 + 3y
Answer:
Step 1: The given equations are 2x + 3y = 8 ...(1) and 2x = 2 + 3y ...(2)
Step 2: From equation (2), divide by 2 to express x in terms of y: x = (2 + 3y) / 2
Step 3: Substitute this value of x into equation (1): 2[ (2 + 3y) / 2 ] + 3y = 8
Step 4: The 2's cancel out: 2 + 3y + 3y = 8
Step 5: Combine like terms: 2 + 6y = 8
Step 6: 6y = 8 - 2
Step 7: 6y = 6
Step 8: y = 1
Step 9: Substitute y = 1 back into the expression for x: x = (2 + 3(1)) / 2
Step 10: x = 5 / 2 = 2.5
Step 11: The solution is x = 2.5 and y = 1.
3. 0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116
Answer:
Step 1: The given equations are 0.2x + 0.1y = 25 ...(1) and 2(x - 2) - 1.6y = 116 ...(2)
Step 2: Simplify equation (2): 2x - 4 - 1.6y = 116
Step 3: 2x - 1.6y = 116 + 4 = 120
Step 4: From equation (1), let's express y in terms of x. Subtract 0.2x from both sides: 0.1y = 25 - 0.2x
Step 5: Multiply by 10 to isolate y: y = 250 - 2x
Step 6: Substitute this y into the simplified equation (2): 2x - 1.6(250 - 2x) = 120
Step 7: Expand: 2x - 400 + 3.2x = 120
Step 8: Combine like terms: 5.2x - 400 = 120
Step 9: 5.2x = 120 + 400 = 520
Step 10: x = 520 / 5.2 = 100
Step 11: Substitute x = 100 into y = 250 - 2x: y = 250 - 2(100)
Step 12: y = 250 - 200 = 50
Step 13: The solution is x = 100 and y = 50.
4. 6x = 7y + 7
7y - x = 8
Answer:
Step 1: The given equations are 6x = 7y + 7 ...(1) and 7y - x = 8 ...(2)
Step 2: From equation (2), let's isolate x: x = 7y - 8
Step 3: Substitute this expression for x into equation (1): 6(7y - 8) = 7y + 7
Step 4: Expand the brackets: 42y - 48 = 7y + 7
Step 5: Move all y terms to one side: 42y - 7y = 7 + 48
Step 6: Simplify: 35y = 55
Step 7: y = 55 / 35
Step 8: Reduce the fraction by dividing by 5: y = 11 / 7
Step 9: Substitute y = 11/7 back into x = 7y - 8
Step 10: x = 7(11 / 7) - 8
Step 11: The 7's cancel out: x = 11 - 8 = 3
Step 12: The solution is x = 3 and y = 11/7.
5. y = 4x - 7
16x - 5y = 25
Answer:
Step 1: The given equations are y = 4x - 7 ...(1) and 16x - 5y = 25 ...(2)
Step 2: Equation (1) already gives y in terms of x. We can directly substitute it into equation (2).
Step 3: 16x - 5(4x - 7) = 25
Step 4: Expand the brackets: 16x - 20x + 35 = 25
Step 5: Combine x terms: -4x + 35 = 25
Step 6: Move 35 to the other side: -4x = 25 - 35
Step 7: -4x = -10
Step 8: x = -10 / -4 = 2.5
Step 9: Substitute x = 2.5 into equation (1): y = 4(2.5) - 7
Step 10: y = 10 - 7 = 3
Step 11: The solution is x = 2.5 and y = 3.
6. 1.5x + 0.1y = 6.2
3x - 0.4y = 11.2
Answer:
Step 1: The given equations are 1.5x + 0.1y = 6.2 ...(1) and 3x - 0.4y = 11.2 ...(2)
Step 2: Multiply both equations by 10 to remove decimals for easier calculation.
Step 3: New Equation (1): 15x + y = 62
Step 4: New Equation (2): 30x - 4y = 112
Step 5: From New Equation (1), express y in terms of x: y = 62 - 15x
Step 6: Substitute this into New Equation (2): 30x - 4(62 - 15x) = 112
Step 7: Expand the brackets: 30x - 248 + 60x = 112
Step 8: Combine like terms: 90x - 248 = 112
Step 9: 90x = 112 + 248
Step 10: 90x = 360
Step 11: x = 360 / 90 = 4
Step 12: Substitute x = 4 into y = 62 - 15x: y = 62 - 15(4)
Step 13: y = 62 - 60 = 2
Step 14: The solution is x = 4 and y = 2.
7. 2(x - 3) + 3(y - 5) = 0
5(x - 1) + 4(y - 4) = 0
Answer:
Step 1: Expand the given equations to simplify them.
Step 2: Equation (1): 2x - 6 + 3y - 15 = 0 => 2x + 3y - 21 = 0 => 2x + 3y = 21
Step 3: Equation (2): 5x - 5 + 4y - 16 = 0 => 5x + 4y - 21 = 0 => 5x + 4y = 21
Step 4: From simplified equation (1), express x in terms of y: 2x = 21 - 3y => x = (21 - 3y) / 2
Step 5: Substitute this into simplified equation (2): 5[ (21 - 3y) / 2 ] + 4y = 21
Step 6: Multiply the entire equation by 2 to remove the fraction: 5(21 - 3y) + 8y = 42
Step 7: Expand: 105 - 15y + 8y = 42
Step 8: Combine y terms: 105 - 7y = 42
Step 9: -7y = 42 - 105
Step 10: -7y = -63
Step 11: y = -63 / -7 = 9
Step 12: Substitute y = 9 back into the expression for x: x = (21 - 3(9)) / 2
Step 13: x = (21 - 27) / 2 = -6 / 2 = -3
Step 14: The solution is x = -3 and y = 9.
8. (2x+1)/7 + (5y-3)/3 = 12
(3x+2)/2 - (4y+3)/9 = 13
Answer:
Step 1: Simplify equation (1) by multiplying it entirely by the LCM of denominators (21).
Step 2: 3(2x + 1) + 7(5y - 3) = 12 * 21
Step 3: 6x + 3 + 35y - 21 = 252
Step 4: 6x + 35y - 18 = 252 => 6x + 35y = 270 ...(Simplified Eq 1)
Step 5: Simplify equation (2) by multiplying it entirely by the LCM of denominators (18).
Step 6: 9(3x + 2) - 2(4y + 3) = 13 * 18
Step 7: 27x + 18 - 8y - 6 = 234
Step 8: 27x - 8y + 12 = 234 => 27x - 8y = 222 ...(Simplified Eq 2)
Step 9: From Simplified Eq 1, express x in terms of y: x = (270 - 35y) / 6
Step 10: Substitute this into Simplified Eq 2: 27[ (270 - 35y) / 6 ] - 8y = 222
Step 11: Simplify the fraction by dividing 27 and 6 by 3: 9(270 - 35y) / 2 - 8y = 222
Step 12: Multiply the entire equation by 2: 9(270 - 35y) - 16y = 444
Step 13: Expand: 2430 - 315y - 16y = 444
Step 14: -331y = 444 - 2430
Step 15: -331y = -1986
Step 16: y = -1986 / -331 = 6
Step 17: Substitute y = 6 into the expression for x: x = (270 - 35(6)) / 6
Step 18: x = (270 - 210) / 6 = 60 / 6 = 10
Step 19: The solution is x = 10 and y = 6.
For solving each pair of equations, use the method of elimination by equating coefficients :
9. 3x - y = 23
x/3 + y/4 = 4
Answer:
Step 1: The given equations are 3x - y = 23 ...(1) and x/3 + y/4 = 4 ...(2)
Step 2: Simplify equation (2) by multiplying by the LCM, which is 12: 4x + 3y = 48 ...(3)
Step 3: To equate the coefficients of y, multiply equation (1) by 3: 9x - 3y = 69 ...(4)
Step 4: Add equation (3) and equation (4): (4x + 3y) + (9x - 3y) = 48 + 69
Step 5: The y terms cancel out: 13x = 117
Step 6: x = 117 / 13 = 9
Step 7: Substitute x = 9 back into equation (1): 3(9) - y = 23
Step 8: 27 - y = 23
Step 9: y = 27 - 23 = 4
Step 10: The solution is x = 9 and y = 4.
10. 5y/2 - x/3 = 8
y/2 + 5x/3 = 12
Answer:
Step 1: Simplify both given equations by multiplying them by the LCM (which is 6) to remove fractions.
Step 2: Equation (1) becomes: 3(5y) - 2(x) = 48 => 15y - 2x = 48 => -2x + 15y = 48 ...(3)
Step 3: Equation (2) becomes: 3(y) + 2(5x) = 72 => 3y + 10x = 72 => 10x + 3y = 72 ...(4)
Step 4: To equate the coefficients of x, multiply equation (3) by 5: -10x + 75y = 240 ...(5)
Step 5: Add equation (4) and equation (5): (10x + 3y) + (-10x + 75y) = 72 + 240
Step 6: The x terms cancel out: 78y = 312
Step 7: y = 312 / 78 = 4
Step 8: Substitute y = 4 into equation (4): 10x + 3(4) = 72
Step 9: 10x + 12 = 72
Step 10: 10x = 60 => x = 6
Step 11: The solution is x = 6 and y = 4.
11. 1/5(x - 2) = 1/4(1 - y)
26x + 3y + 4 = 0
Answer:
Step 1: Simplify the first equation by cross-multiplying: 4(x - 2) = 5(1 - y)
Step 2: Expand: 4x - 8 = 5 - 5y
Step 3: Rearrange into standard form: 4x + 5y = 13 ...(1)
Step 4: Rearrange the second equation: 26x + 3y = -4 ...(2)
Step 5: To equate coefficients of y, multiply equation (1) by 3 and equation (2) by 5.
Step 6: New equation (1): 12x + 15y = 39 ...(3)
Step 7: New equation (2): 130x + 15y = -20 ...(4)
Step 8: Subtract equation (3) from equation (4): (130x - 12x) + (15y - 15y) = -20 - 39
Step 9: 118x = -59
Step 10: x = -59 / 118 = -1/2 (or -0.5)
Step 11: Substitute x = -1/2 into equation (1): 4(-1/2) + 5y = 13
Step 12: -2 + 5y = 13
Step 13: 5y = 15 => y = 3
Step 14: The solution is x = -1/2 and y = 3.
12. (x-y)/6 = 2(4-x)
2x + y = 3(x-4)
Answer:
Step 1: Simplify the first equation: x - y = 12(4 - x)
Step 2: x - y = 48 - 12x
Step 3: Rearrange: 13x - y = 48 ...(1)
Step 4: Simplify the second equation: 2x + y = 3x - 12
Step 5: Rearrange: -x + y = -12 ...(2)
Step 6: The coefficients of y are already equal and opposite (-1 and 1). Add equation (1) and (2).
Step 7: (13x - y) + (-x + y) = 48 + (-12)
Step 8: 12x = 36
Step 9: x = 36 / 12 = 3
Step 10: Substitute x = 3 into equation (2): -3 + y = -12
Step 11: y = -12 + 3 = -9
Step 12: The solution is x = 3 and y = -9.
13. 2x - 3y - 3 = 0
2x/3 + 4y + 1/2 = 0
Answer:
Step 1: Write equation (1) as: 2x - 3y = 3 ...(1)
Step 2: Simplify equation (2) by multiplying by the LCM (6) to remove fractions: 4x + 24y + 3 = 0
Step 3: Rearrange: 4x + 24y = -3 ...(2)
Step 4: To equate the coefficients of x, multiply equation (1) by 2: 4x - 6y = 6 ...(3)
Step 5: Subtract equation (3) from equation (2): (4x - 4x) + (24y - (-6y)) = -3 - 6
Step 6: 30y = -9
Step 7: y = -9 / 30 = -3/10
Step 8: Substitute y = -3/10 into equation (1): 2x - 3(-3/10) = 3
Step 9: 2x + 9/10 = 3
Step 10: 2x = 3 - 9/10 = (30 - 9)/10 = 21/10
Step 11: x = (21/10) / 2 = 21/20
Step 12: The solution is x = 21/20 and y = -3/10.
14. 13x + 11y = 70
11x + 13y = 74
Answer:
Step 1: Notice that the coefficients of x and y are interchanged in these two equations.
Step 2: Add the two equations together: (13x + 11x) + (11y + 13y) = 70 + 74
Step 3: 24x + 24y = 144
Step 4: Divide the entire equation by 24: x + y = 6 ...(A)
Step 5: Now, subtract the second equation from the first: (13x - 11x) + (11y - 13y) = 70 - 74
Step 6: 2x - 2y = -4
Step 7: Divide this equation by 2: x - y = -2 ...(B)
Step 8: Now, solve the simple equations (A) and (B) by adding them: (x + y) + (x - y) = 6 - 2
Step 9: 2x = 4 => x = 2
Step 10: Substitute x = 2 into equation (A): 2 + y = 6 => y = 4
Step 11: The solution is x = 2 and y = 4.
15. 41x + 53y = 135
53x + 41y = 147
Answer:
Step 1: The coefficients of x and y are interchanged. Use the addition and subtraction method.
Step 2: Add the two equations: (41x + 53x) + (53y + 41y) = 135 + 147
Step 3: 94x + 94y = 282
Step 4: Divide the equation by 94: x + y = 3 ...(A)
Step 5: Subtract the first equation from the second: (53x - 41x) + (41y - 53y) = 147 - 135
Step 6: 12x - 12y = 12
Step 7: Divide the equation by 12: x - y = 1 ...(B)
Step 8: Add the simplified equations (A) and (B): (x + y) + (x - y) = 3 + 1
Step 9: 2x = 4 => x = 2
Step 10: Substitute x = 2 into equation (A): 2 + y = 3 => y = 1
Step 11: The solution is x = 2 and y = 1.
16. If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x. (HOTS)
Answer:
Step 1: The given equations are 2x + y = 23 ...(1) and 4x - y = 19 ...(2)
Step 2: Add the two equations to eliminate y: (2x + 4x) + (y - y) = 23 + 19
Step 3: 6x = 42
Step 4: x = 42 / 6 = 7
Step 5: Substitute x = 7 into equation (1): 2(7) + y = 23
Step 6: 14 + y = 23
Step 7: y = 23 - 14 = 9
Step 8: Now, find the value of (x - 3y). Substitute x = 7 and y = 9.
Step 9: x - 3y = 7 - 3(9) = 7 - 27 = -20
Step 10: Now, find the value of (5y - 2x). Substitute x = 7 and y = 9.
Step 11: 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31
Step 12: The final values are x - 3y = -20 and 5y - 2x = 31.
17. If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.
Answer:
Step 1: Rearrange the first equation into standard form: 7x - 10y = 4 ...(1)
Step 2: The second equation is 12x + 18y = 1 ...(2)
Step 3: Multiply equation (1) by 9 and equation (2) by 5 to equate coefficients of y.
Step 4: 63x - 90y = 36 ...(3)
Step 5: 60x + 90y = 5 ...(4)
Step 6: Add equation (3) and (4): 123x = 41
Step 7: x = 41 / 123 = 1/3
Step 8: Substitute x = 1/3 into equation (2): 12(1/3) + 18y = 1
Step 9: 4 + 18y = 1 => 18y = -3
Step 10: y = -3 / 18 = -1/6
Step 11: Now find the value of (4x + 6y) by substituting x = 1/3 and y = -1/6.
Step 12: 4(1/3) + 6(-1/6) = 4/3 - 1 = 4/3 - 3/3 = 1/3
Step 13: Now find the value of (8y - x) by substituting x = 1/3 and y = -1/6.
Step 14: 8(-1/6) - 1/3 = -4/3 - 1/3 = -5/3
Step 15: The final values are 4x + 6y = 1/3 and 8y - x = -5/3.
18. Solve for x and y :
(i) (y+7)/5 = (2y-x)/4 + 3x - 5; (7-5x)/2 + (3-4y)/6 = 5y - 18
Answer:
Step 1: Simplify the first equation by taking the LCM (4) on the right hand side: (y+7)/5 = (2y - x + 12x - 20)/4
Step 2: Cross-multiply: 4(y + 7) = 5(11x + 2y - 20)
Step 3: Expand: 4y + 28 = 55x + 10y - 100
Step 4: Group variables on one side: 55x + 6y = 128 ...(1)
Step 5: Simplify the second equation by multiplying the entire equation by the LCM (6).
Step 6: 3(7 - 5x) + (3 - 4y) = 6(5y - 18)
Step 7: Expand: 21 - 15x + 3 - 4y = 30y - 108
Step 8: Group variables on one side: 15x + 34y = 132 ...(2)
Step 9: To equate coefficients of x, multiply equation (1) by 3 and equation (2) by 11.
Step 10: 165x + 18y = 384 ...(3)
Step 11: 165x + 374y = 1452 ...(4)
Step 12: Subtract equation (3) from equation (4): 356y = 1068
Step 13: y = 1068 / 356 = 3
Step 14: Substitute y = 3 into equation (1): 55x + 6(3) = 128
Step 15: 55x + 18 = 128 => 55x = 110 => x = 2
Step 16: The solution is x = 2 and y = 3.
(ii) 4x = 17 - (x-y)/8; 2y + x = 2 + (5y+2)/3
Answer:
Step 1: Simplify the first equation by multiplying by 8 to remove the fraction: 32x = 136 - (x - y)
Step 2: Expand and rearrange: 32x = 136 - x + y => 33x - y = 136 ...(1)
Step 3: Simplify the second equation by multiplying by 3 to remove the fraction: 3(2y + x) = 3(2) + (5y + 2)
Step 4: Expand: 6y + 3x = 6 + 5y + 2
Step 5: Rearrange: 3x + y = 8 ...(2)
Step 6: Add equation (1) and equation (2) to eliminate y: (33x - y) + (3x + y) = 136 + 8
Step 7: 36x = 144
Step 8: x = 144 / 36 = 4
Step 9: Substitute x = 4 into equation (2): 3(4) + y = 8
Step 10: 12 + y = 8 => y = 8 - 12 = -4
Step 11: The solution is x = 4 and y = -4.
EXERCISE 6 (B)
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) Solution of equations x + y = 8 and 3x - 5y = 0 is :
(i) x = 5, y = -3
(ii) x = 5, y = 3
(iii) x = -5, y = 3
(iv) x = -5, y = -3
Answer: (ii)
Step 1: From the first equation x + y = 8, express x as x = 8 - y.
Step 2: Substitute this into the second equation: 3(8 - y) - 5y = 0
Step 3: 24 - 3y - 5y = 0
Step 4: 24 = 8y => y = 3
Step 5: Substitute y = 3 back into x = 8 - y: x = 8 - 3 = 5
Step 6: The solution is x = 5, y = 3. This matches option (ii).
(b) Solution of equations (4a-3)/6 + (5b-7)/2 = 18 - 5a and a + b = 5 is :
(i) a = -3, b = 2
(ii) a = -3, b = -2
(iii) a = 3, b = 2
(iv) a = 3, b = -2
Answer: (iii)
Step 1: From a + b = 5, express b as b = 5 - a.
Step 2: Substitute b = 5 - a into the first equation: (4a - 3)/6 + (5(5 - a) - 7)/2 = 18 - 5a
Step 3: Simplify the numerator: (4a - 3)/6 + (25 - 5a - 7)/2 = 18 - 5a
Step 4: (4a - 3)/6 + (18 - 5a)/2 = 18 - 5a
Step 5: Multiply the entire equation by 6: (4a - 3) + 3(18 - 5a) = 6(18 - 5a)
Step 6: 4a - 3 + 54 - 15a = 108 - 30a
Step 7: -11a + 51 = 108 - 30a
Step 8: 19a = 57 => a = 3
Step 9: Substitute a = 3 into b = 5 - a: b = 5 - 3 = 2
Step 10: The solution is a = 3, b = 2. This matches option (iii).
(c) Solution of equations x/7 - y/2 = 2 and (x-y)/3 = 3 is :
(i) x = 7, y = 2
(ii) x = -7, y = 2
(iii) x = -7, y = -2
(iv) x = 7, y = -2
Answer: (iv)
Step 1: From the second equation (x-y)/3 = 3, multiply by 3: x - y = 9 => x = y + 9
Step 2: Substitute x = y + 9 into the first equation: (y + 9)/7 - y/2 = 2
Step 3: Multiply the entire equation by LCM 14: 2(y + 9) - 7y = 28
Step 4: 2y + 18 - 7y = 28
Step 5: -5y + 18 = 28
Step 6: -5y = 10 => y = -2
Step 7: Substitute y = -2 into x = y + 9: x = -2 + 9 = 7
Step 8: The solution is x = 7, y = -2. This matches option (iv).
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Question
What is the general form of a linear equation in two variables?
Answer
$ax + by + c = 0$, where $a, b,$ and $c$ are real numbers and $x, y$ are variables.
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Each variable must have a degree of $1$ (one).
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Answer
simultaneous (linear)
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To find the specific values of the variables that satisfy both equations at the same time.
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What are the three primary algebraic methods discussed for solving simultaneous equations?
Answer
1. Elimination by substitution, 2. Elimination by equating coefficients, and 3. Cross-multiplication.
Question
Method of elimination by substitution: What is the objective of Step 1?
Answer
To express one variable in terms of the other variable using one of the given equations.
Question
Method of elimination by substitution: What is done with the expression obtained in Step 1?
Answer
It is substituted into the other equation to create a single-variable equation.
Question
Method of elimination by substitution: How is the value of the second unknown variable found in the final step?
Answer
By substituting the solved value of the first variable back into the expression from Step 1.
Question
Method of elimination by equating coefficients: What is the first step to prepare the equations?
Answer
Multiply one or both equations by suitable numbers so the coefficients of one variable become numerically equal.
Question
Method of equating coefficients: Under what condition should you subtract one equation from the other?
Answer
When the terms with equal numerical coefficients have the same sign.
Question
Method of equating coefficients: Under what condition should you add the two equations?
Answer
When the terms with equal numerical coefficients have opposite signs.
Question
In equations where the $x$-coefficient of the first is the $y$-coefficient of the second (and vice-versa), what is the first simplifying step?
Answer
Add the two equations together to obtain a simpler equation, usually in the form $x + y = k$.
Question
In equations where the $x$-coefficient of the first is the $y$-coefficient of the second (and vice-versa), what is the second simplifying step?
Answer
Subtract one equation from the other to obtain a simpler equation, usually in the form $x - y = k$.
Question
To solve equations using cross-multiplication, in what standard form must the equations first be expressed?
Answer
$a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.
Question
In the method of cross-multiplication, what is the sequence of coefficients and constants used for the arrow arrangement?
Answer
$b, c, a, b$ (the coefficients of $y$, then constants, then coefficients of $x$, then coefficients of $y$ again).
Question
Cross-multiplication rule: When calculating products between terms connected by arrows, which product is subtracted?
Answer
The product of the numbers connected by the upward arrow.
Question
Cross-multiplication rule: When calculating products between terms connected by arrows, which product is calculated first?
Answer
The product of the numbers connected by the downward arrow.
Question
What is the formula for the variable $x$ in the method of cross-multiplication?
Answer
$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$
Question
What is the formula for the variable $y$ in the method of cross-multiplication?
Answer
$y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$
Question
In the cross-multiplication formulas for $x$ and $y$, what is the common denominator?
Answer
$a_1b_2 - a_2b_1$
Question
How should you solve a system where the variables are only in the denominators, such as $\frac{7}{x} + \frac{8}{y} = 2$?
Answer
Use substitutions such as $a = \frac{1}{x}$ and $b = \frac{1}{y}$ to transform them into linear equations.
Question
For complex denominators like $\frac{30}{x-y} + \frac{44}{x+y} = 10$, what substitution is recommended for $x-y$?
Answer
Let $x - y = a$ (or $\frac{1}{x-y} = a$).
Question
What is the first step in solving a word problem using simultaneous equations?
Answer
Assume the two unknown quantities to be the variables $x$ and $y$.
Question
In problems based on numbers, if the sum of two numbers is $12$ and their difference is $2$, what are the resulting equations?
Answer
$x + y = 12$ and $x - y = 2$.
Question
In fraction problems, how is a fraction typically represented algebraically?
Answer
As $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator.
Question
In fraction problems, if the numerator is decreased by $1$, what is the new expression for the numerator?
Answer
$x - 1$
Question
In fraction problems, if the denominator is increased by $5$, what is the new expression for the denominator?
Answer
$y + 5$
Question
How is a two-digit number represented algebraically if $x$ is the tens digit and $y$ is the units digit?
Answer
$10x + y$
Question
How is a two-digit number expressed algebraically after its digits are reversed?
Answer
$10y + x$
Question
If the sum of the digits of a two-digit number is $7$, what is the algebraic equation?
Answer
$x + y = 7$
Question
If a person's present age is $x$ years, what will their age be '$n$' years hence (in the future)?
Answer
$x + n$
Question
If a person's present age is $x$ years, what was their age '$n$' years ago?
Answer
$x - n$
Question
In problems involving Cost Price (C.P.), if the C.P. is $x$ and there is a profit of $25\%$, what is the Selling Price (S.P.)?
Answer
$S.P. = x + 25\% \text{ of } x = \frac{5x}{4}$ (or $1.25x$).
Question
In problems involving Cost Price (C.P.), if the C.P. is $y$ and there is a profit of $10\%$, what is the Selling Price (S.P.)?
Answer
$S.P. = y + 10\% \text{ of } y = \frac{11y}{10}$ (or $1.1y$).
Question
In Time and Work problems, if A can finish a piece of work in $x$ days, what is A's work in one day?
Answer
$\frac{1}{x}$
Question
If A and B together can do a piece of work in $15$ days, what equation represents their combined daily work?
Answer
$\frac{1}{x} + \frac{1}{y} = \frac{1}{15}$, where $x$ and $y$ are the days each takes alone.
Question
In problems based on rectangles, if the length is $x$ and the breadth is $y$, what is the area?
Answer
$xy$
Question
In rectangle problems, if the length is reduced by $5$ units and the breadth is increased by $3$ units, what is the expression for the new area?
Answer
$(x - 5)(y + 3)$
Question
If the digits of a two-digit number differ by $3$, what are the two possible algebraic relationships between the tens digit ($x$) and units digit ($y$)?
Answer
$x - y = 3$ or $y - x = 3$.
Question
For equations like $ax + by = c$, what is the term for $c$?
Answer
The constant term.
Question
In taxi fare problems, if there is a fixed charge $x$ and a per kilometre charge $y$, what is the total fare for a distance of $10$ km?
Answer
$x + 10y$
Question
In problems involving mixtures, if $x$ litres of a $90\%$ solution and $y$ litres of a $97\%$ solution result in $21$ litres of a $95\%$ solution, what is the quantity equation?
Answer
$x + y = 21$
Question
In problems involving mixtures, if $x$ litres of a $90\%$ solution and $y$ litres of a $97\%$ solution result in $21$ litres of a $95\%$ solution, what is the concentration equation?
Answer
$90\% \text{ of } x + 97\% \text{ of } y = 95\% \text{ of } 21$.
Question
What is the algebraic representation of 'the sum of the reciprocals of two numbers $x$ and $y$ is $\frac{4}{21}$'?
Answer
$\frac{1}{x} + \frac{1}{y} = \frac{4}{21}$
Question
If the ratio of the present ages of A and B is $5:3$, how can this be written as an equation using variables $x$ and $y$?
Answer
$\frac{x}{y} = \frac{5}{3}$ or $3x - 5y = 0$.
Question
To verify if a pair of values $(x, y)$ is a solution to a system, what must happen when they are substituted into the equations?
Answer
The values must satisfy both equations (the left-hand side must equal the right-hand side in both).
Question
If an equation is presented as $\frac{x+y}{xy} = 2$, how can it be rewritten in a form that allows for substitution?
Answer
$\frac{x}{xy} + \frac{y}{xy} = 2$, which simplifies to $\frac{1}{y} + \frac{1}{x} = 2$.
Question
In two-digit number problems, if reversing the digits increases the number, is the units digit ($y$) or tens digit ($x$) larger?
Answer
The units digit ($y$) is larger ($y > x$).
Question
In age problems, if a father's age is $x$ and the sum of his two children's ages is $y$, what will the sum of the children's ages be in $20$ years?
Answer
$y + 40$ (because each child's age increases by $20$ years).
Question
When solving simultaneous equations with variables in denominators, is it necessary to find the L.C.M. of the denominators?
Answer
No, it is usually simpler to solve without taking the L.C.M. by using substitution or equating coefficients directly.
Question
In a library charge problem where the first three days have a fixed charge ($x$) and subsequent days have a daily charge ($y$), what is the total charge for $7$ days?
Answer
$x + 4y$
Question
For the system $x + y = 3.3$ and $0.6/(3x-2y) = -1$, what is the first step to simplify the second equation?
Answer
Cross-multiply to get $0.6 = -1(3x - 2y)$, which becomes $3x - 2y = -0.6$.
Question
If $10y = 7x - 4$ and $12x + 18y = 1$, what method is easiest if you want to avoid fractions in the initial steps?
Answer
Method of elimination by equating coefficients.
Question
In rectangle area problems, if length is $x$ and breadth is $y$, what equation is formed if area increases by $67$ when length increases by $3$ and breadth increases by $2$?
Answer
$(x + 3)(y + 2) = xy + 67$
Question
In the cross-multiplication arrow diagram, what does the '1' represent in the sequence $\frac{x}{\dots} = \frac{y}{\dots} = \frac{1}{\dots}$?
Answer
The constant terms in the equations when they are on the left-hand side ($= 0$).