Long Question (Question 10)

Posted on May 17, 2020 by Myhometuition

Question 10 (5 marks):
Solve the following simultaneous equations:
x – 3y = 1,
x²+ 3xy + 9y²= 7

Solution:

\begin{array}{l} x - 3 y = 1................... (1) \\ x^{2} + 3 x y + 9 y^{2} = 7................... (2) \\ From (1) : x = 3 y + 1................... (3) \\ Substitute (3) into (2) : \\ {(3 y + 1)}^{2} + 3 (3 y + 1) y + 9 y^{2} = 7 \\ 9 y^{2} + 6 y + 1 + 9 y^{2} + 3 y + 9 y^{2} - 7 = 0 \\ 27 y^{2} + 9 y - 6 = 0 \\ 9 y^{2} + 3 y - 2 = 0 \\ (3 y - 1) (3 y + 2) = 0 \\ y = \frac{1}{3} or - \frac{2}{3} \\ Substitute y into (3) : \\ When y = \frac{1}{3} \\ x = 3 (\frac{1}{3}) + 1 = 2 \\ When y = - \frac{2}{3} \\ x = 3 (- \frac{2}{3}) + 1 = - 1 \\ Hence, the solutions are x = 2, y = \frac{1}{3} or x = - 1, y = - \frac{2}{3} . \end{array}

Long Question (Question 9)

Posted on May 17, 2020 by Myhometuition

Question 9 (7 marks):
Diagram shows the plan of a rectangular garden ABCD. The garden consists of a semicircular pond ATD and grassy area ABCDT.

It is given that DC = 6y metre and BC = 7x metre, x ≠ y. The area of the rectangular garden ABCD is 168 metre² and the perimeter of the grassy area is 60 metre. The pond with uniform depth contains 15.4 metre³ of water.
By using

π = \frac{22}{7}

, find the depth, in metre, of water in the pond.

Solution:

\begin{array}{l} Area of A B C D = 168 \\ (6 y) (7 x) = 168 \\ 42 x y = 168 \\ x y = 4................... (1) \\ Perimeter of grassy area = 60 \\ 6 y + 6 y + 7 x + (\frac{1}{2} \times 2 \times \frac{22^{11}}{7} \times \frac{7 x}{2}) = 60 \\ 12 y + 18 x = 60 \\ 2 y + 3 x = 10................... (2) \\ From (1) : x y = 4 \\ x = \frac{4}{y} ................... (3) \\ Substitute (3) into (2): \\ 2 y + 3 (\frac{4}{y}) = 10 \\ 2 y^{2} + 12 = 10 y \\ y^{2} - 5 y + 6 = 0 \\ (y - 2) (y - 3) = 0 \\ y - 2 = 0 \\ y = 2 \\ or \\ y - 3 = 0 \\ y = 3 \\ Substitute the values of y into (3): \\ When y = 2 \\ x = \frac{4}{2} = 2 (ignored, x \neq y) \\ When y = 3 \\ x = \frac{4}{3} \\ Given volume of water = 15.4 \\ \frac{1}{2} (\frac{22}{7}) {(\frac{7 x}{2})}^{2} d = 15.4 \\ \frac{1}{2} (\frac{22}{7}) {[\frac{7 (\frac{4}{3})}{2}]}^{2} d = 15.4 \\ \frac{11}{7} {(\frac{14}{3})}^{2} d = 15.4 \\ \frac{308}{9} d = 15.4 \\ d = 0.45 m \\ Thus, the depth of water in the pond is 0 .45 m . \end{array}

4.1 Simultaneous Equations

Posted on April 18, 2020 by user

4.1 Simultaneous Equations

(A) Steps in solving simultaneous equations:

For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
Substitute the linear equation into the non-linear equation.
Simplify and expressed the equation in the general form of quadratic equation $a x^{2} + b x + c = 0$
Solve the quadratic equation.
Find the value of the second unknown by substituting the value obtained into the linear equation.

Example:
Solve the following simultaneous equations.

\begin{array}{l} y + x = 9 \\ x y = 20 \end{array}

Solution:
For the linear equation, arrange so that one of the unknown becomes the subject of the equation.

\begin{array}{l} y + x = 9 \\ y = 9 - x \end{array}

Substitute the linear equation into the non-linear equation.

\begin{array}{l} x y = 20 \\ x (9 - x) = 20 \\ 9 x - x^{2} = 20 \end{array}

Simplify and expressed the equation in the general form of quadratic equation

a x^{2} + b x + c = 0

\begin{array}{l} 9 x - x^{2} = 20 \\ x^{2} - 9 x + 20 = 0 \end{array}

Solve the quadratic equation.

\begin{array}{l} x^{2} - 9 x + 20 = 0 \\ (x - 4) (x - 5) = 0 \\ x = 4 or x = 5 \end{array}

Find the value of the second unknown by substituting the value obtained into the linear equation.

\begin{array}{l} When x = 4, \\ y = 9 - x = 9 - 4 = 5 \\ When x = 5, \\ y = 9 - x = 9 - 5 = 4 \end{array}

Long Questions (Question 7 & 8)

Posted on April 18, 2020 by user

4.2.4 Simultaneous Equations, Long Questions

Question 7:

Solve the following simultaneous equations.

5y – 6x= 2

\frac{4 y}{x} - \frac{3 x}{y} = 4.

Solution:

5y – 6x= 2 ----- (1)

\begin{array}{l} \frac{4 y}{x} - \frac{3 x}{y} = 4 ------- (2) \\ From (1), \\ y = \frac{2 + 6 x}{5} \end{array}

Substitute (3) into (2),

\begin{array}{l} \frac{4 (\frac{2 + 6 x}{5})}{x} - \frac{3 x}{(\frac{2 + 6 x}{5})} = 4 \\ \frac{8 + 24 x}{5 x} - \frac{15 x}{2 + 6 x} = 4 \\ \frac{(8 + 24 x) (2 + 6 x) - (15 x) (5 x)}{5 x (2 + 6 x)} = 4 \end{array}

16 + 48x + 48x + 144x² – 75x² = 20x (2 + 6x)

69x² + 96x + 16 = 40x + 120x²

51x² – 56x – 16 = 0

(3x – 4)(17x + 4) = 0

3x – 4 = 0 or 17x + 4 = 0

x = \frac{4}{3} or x = - \frac{4}{17}

\begin{array}{l} Substitute x = \frac{4}{3} into (3), \\ y = \frac{2 + 6 (\frac{4}{3})}{5} = 2 \end{array}

\begin{array}{l} Substitute x = - \frac{4}{17} into (3), \\ y = \frac{2 + 6 (- \frac{4}{17})}{5} = \frac{2}{17} \end{array}

∴ The solutions are (\frac{4}{3}, 2) and (- \frac{4}{17}, \frac{2}{17}) .

Question 8:
Solve the following simultaneous equations.

\begin{array}{l} x + 2 y = 1 \\ 2 x^{2} + y^{2} + x y = 5 \end{array}

Give your answer correct to three decimal places.

Solution:

\begin{array}{l} x + 2 y = 1.......... (1) \\ 2 x^{2} + y^{2} + x y = 5.......... (2) \\ x = 1 - 2 y .......... (3) \\ Substitute (3) into (2), \\ 2 {(1 - 2 y)}^{2} + y^{2} + (1 - 2 y) y = 5 \\ 2 (1 - 4 y + 4 y^{2}) + y^{2} + y - 2 y^{2} = 5 \\ 2 - 8 y + 8 y^{2} - y^{2} + y - 5 = 0 \\ 7 y^{2} - 7 y - 3 = 0 \\ a = 7, b = - 7, c = - 3 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ y = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (7) (- 3)}}{2 (7)} \\ y = \frac{7 \pm \sqrt{133}}{14} \\ y = 1.324 or - 0.324 \\ From x = 1 - 2 y \\ When y = 1.324, \\ x = 1 - 2 (1.324) = 1.648 \\ When y = - 0.324, \\ x = 1 - 2 (- 0.324) = 1.648 \\ ∴ The solutions are (1.648, 1.324) and (1.648, - 0.324) . \end{array}

Long Questions (Question 5 & 6)

Posted on April 18, 2020 by user

Question 5:

Lisa has a rectangular plot of land. She plants orchid and rears fish in the areas as shown on diagram above. The area used for planting orchid is 460 m² and the perimeter of the rectangular fish pond is 48 m. Find the value of x and y.

Solution:

Area used for planting orchid = 460 m²

10 (30 – y) + xy = 460

300 – 10y + xy = 460

xy – 10y = 160

y (x – 10) = 160

y = \frac{160}{x - 10} ------- (1)

Perimeter of the rectangular fish pond = 48 m

2 (x – 10) + 2 (30 – y) = 48

2x – 20 + 60 – 2y = 48

2x – 2y= 8
x – y = 4

x = 4 + y ------ (2)

Substitute (2) into (1):

\begin{array}{l} y = \frac{160}{x - 10} \\ y = \frac{160}{4 + y - 10} \\ y = \frac{160}{y - 6} \\ y^{2} - 6 y - 160 = 0 \\ (y - 16) (y + 10) = 0 \\ y = 16 or y = - 10 (not accepted) \end{array}

From (2),

When y = 16

x = 4 + 16 = 20

Question 6:

In the diagram above, PQRS is a rectangular piece of paper with an area of 112 cm². A semicircle STR is cut from this piece of paper. The perimeter of the remaining piece of paper is 52 cm. Using

π = \frac{22}{7}

, find the integer value of x and y.

Solution:

Area of the rectangle PQRS = 112 cm²

Therefore, (14x)(2y) = 112

28xy= 112

xy = 4 ------ (1)

Perimeter of PSTRQ = 52 cm

PS + QR + PQ + Length of arc STR = 52

2y + 2y + 14x + ½ (2πr) = 52

4y + 14x +

(\frac{22}{7}) (7 x)

= 52

4y + 14x + 22x = 52

4y + 36x = 52

y + 9x = 13 ------ (2)

From equation (2) : y = 13 – 9x ------ (3)

Substitute (3) into (1) :

x (13 – 9x) = 4

13x – 9x² = 4

9x² – 13x + 4 = 0

(x – 1)(9x – 4) = 0

x = 1 or

\frac{4}{9} (not an interger, therefore not accepted)

From (3) :

When x = 1,

y = 13 – 9(1) = 4.

Long Questions (Question 3 & 4)

Posted on April 18, 2020 by user

Question 3:

Solve the following simultaneous equations.

3y – 2x= – 4

y² + 4x² = 2

Solution:

3y – 2x= – 4 -----(1)

y² + 4x²= 2 -----(2)

\begin{array}{l} From (1), \\ y = \frac{2 x - 4}{3} ------- (3) \\ Substitute (3) into (2), \\ {(\frac{2 x - 4}{3})}^{2} + 4 x^{2} = 2 \\ (\frac{4 x^{2} - 16 x + 16}{9}) + 4 x^{2} = 2 \\ 4 x^{2} - 16 x + 16 + 36 x^{2} = 18 (\times 9) \\ 40 x^{2} - 16 x - 2 = 0 \\ 20 x^{2} - 8 x - 1 = 0 \\ (10 x + 1) (2 x - 1) = 0 \\ x = - \frac{1}{10} or x = \frac{1}{2} \\ Substitute the values of x into (3), \\ When x = - \frac{1}{10}, \\ y = \frac{2 (- \frac{1}{10}) - 4}{3} = - 1 \frac{2}{5} \\ When x = \frac{1}{2}, \\ y = \frac{2 (\frac{1}{2}) - 4}{3} = - \frac{3}{3} = 1 \\ The solutions are x = - \frac{1}{10}, y = - 1 \frac{2}{5} and x = \frac{1}{2}, y = 1. \end{array}

Question 4:

Solve the simultaneous equations x – 3y = –1 and y + yx – 2x = 0.

Give your answers correct to three decimal places.

Solution:

x – 3y = –1 -----(1)

y + yx – 2x = 0 -----(2)

From (1),

x = 3y – 1 -----(3)

Substitute (3) into (2),

y + y (3y – 1) – 2(3y – 1) = 0

y + 3y² – y – 6y+ 2 = 0

3y² – 6y + 2 = 0

\begin{array}{l} a = 3, b = - 6, c = 2 \\ y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ y = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (3) (2)}}{2 (3)} \\ y = \frac{6 \pm \sqrt{12}}{6} \\ y = 1.577 or 0 .423 \end{array}

Substitute the values of y into (3).

When y = 1.577,

x = 3 (1.577) – 1 = 3.731 (correct to 3 decimal places)

When y = 0.423,

x = 3 (0.423) – 1 = 0.269 (correct to 3 decimal places)

The solutions are x = 3.731, y = 1.577 and x = 0.269, y = 0.423.

Long Questions (Question 1 & 2)

Posted on April 18, 2020 by user

Question 1:

Solve the following simultaneous equations.

\begin{array}{l} y + 2 x = 2 \\ \frac{2}{x} + \frac{1}{y} = 5 \end{array}

Solution:

\begin{array}{l} y + 2 x = 2 - - - - (1) \\ \frac{2}{x} + \frac{1}{y} = 5 - - - - (2) \\ y = 2 - 2 x - - - - (3) \\ substitute (3) into (2), \\ \frac{2}{x} + \frac{1}{2 - 2 x} = 5 \\ \frac{2 (2 - 2 x) + x}{x (2 - 2 x)} = 5 \\ 4 - 4 x + x = 5 x (2 - 2 x) \\ 4 - 3 x = 10 x - 10 x^{2} \\ 10 x^{2} - 13 x + 4 = 0 \\ (5 x - 4) (2 x - 1) = 0 \\ 5 x - 4 = 0 or     2 x - 1 = 0 \\ x = \frac{4}{5} or x = \frac{1}{2} \\ Substitute values of x into (3), \\ When x = \frac{4}{5}, \\ y = 2 - 2 (\frac{4}{5}) = \frac{2}{5} \\ When x = \frac{1}{2} \\ y = 2 - 2 (\frac{1}{2}) = 1 \\ The solutions are x = \frac{4}{5}, y = \frac{2}{5} and x = \frac{1}{2}, y = 1 \end{array}

Question 2:

Solve the following simultaneous equations.

x – 3y + 5 = 3y + 5y²– 6 – x = 0

Solution:

x – 3y + 5 = 0

x = 3y – 5 -----(1)

3y + 5y² – 6 – x = 0 -----(2)

Substitute (1) into (2),

3y + 5y² – 6 – (3y – 5) = 0

3y + 5y² – 6 – 3y + 5 = 0

5y² – 1 = 0

5y² = 1

y = ±0.447

Substitute the values of y into (1),

When y = 0.447

x = 3 (0.447) – 5

x = –3.659

When y = – 0.447

x = 3 (–0.447) – 5

x = –6.341

The solutions are x = –3.659, y = 0.447 and x = –6.341, y = – 0.447.

Simultaneous Equations (Example 1 & 2)

Posted on April 18, 2020 by user

Example 1:

Solve the simultaneous equations.

x + \frac{1}{4} y = 1 and y^{2} - 8 = 4 x .

Solution:

\begin{array}{l} x + \frac{1}{4} y = 1 \to (1) \\ y^{2} - 8 = 4 x \to (2) \\ x = 1 - \frac{1}{4} y \to (3) \end{array}

Substitute (3) into (2),

\begin{array}{l} y^{2} - 8 = 4 (1 - \frac{1}{4} y) \\ y^{2} - 8 = 4 - \frac{4}{4} y \end{array}

y² + y – 12 = 0

(y + 4)(y – 3) = 0

y = –4 or y = 3

Substitute the values of y into (3),

\begin{array}{l} when y = - 4, \\ x = 1 - \frac{1}{4} (- 4) = 2 \\ when y = 3, \\ x = 1 - \frac{1}{4} (3) = \frac{1}{4} \\ The solutions are x = 2, y = - 4 and x = \frac{1}{4}, y = 3 \end{array}

Example 2:

Solve the simultaneous equations 2x + y = 1 and 2x²+ y² + xy = 5.

Correct your answer to three decimal places.

Solution:

2x + y = 1-----(1)

2x² + y²+ xy = 5-----(2)

From (1),

y = 1 – 2x-----(3)

Substitute (3) into (2).

2x² + (1 – 2x)² + x(1 – 2x) = 5

2x² + (1 – 2x)(1 – 2x) + x – 2x² = 5

1 – 2x – 2x + 4x² + x – 5 = 0

4x² – 3x – 4 = 0

\begin{array}{l} From x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ a = 4, b = - 3, c = - 4 \\ x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (4) (- 4)}}{2 (4)} \\ x = \frac{3 \pm \sqrt{73}}{8} \\ x = - 0.693 or 1.443 \end{array}

Substitute the values of x into (3).

When x = –0.693,

y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places)

When x = 1.443,

y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places)

The solutions are x = –0.693, y = 2.386 and x = 1.443, y = –1.886.