Answer :
### (a) Finding the Price of the Cake:
Let's denote the original price of the cake as [tex]\( x \)[/tex] Tsh.
1. Set up the problem:
- Amina has Tsh 1800 to spend.
- If the price decreases by Tsh 20, she can buy 3 more cakes.
- With the original price, the number of cakes she can buy is [tex]\( \frac{1800}{x} \)[/tex].
- With the decreased price [tex]\( x - 20 \)[/tex], the number of cakes she can buy is [tex]\( \frac{1800}{x - 20} \)[/tex].
2. Form the equation:
[tex]\[ \frac{1800}{x - 20} = \frac{1800}{x} + 3 \][/tex]
3. Eliminate the fractions:
Multiply through by [tex]\( x(x - 20) \)[/tex]:
[tex]\[ 1800x = 1800(x - 20) + 3x(x - 20) \][/tex]
4. Expand and simplify:
[tex]\[ 1800x = 1800x - 36000 + 3x^2 - 60x \][/tex]
[tex]\[ 0 = 3x^2 - 60x - 36000 \][/tex]
5. Form the quadratic equation:
[tex]\[ 3x^2 - 60x - 36000 = 0 \][/tex]
6. Solve using the quadratic formula:
The quadratic formula is given by [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -60 \)[/tex], and [tex]\( c = -36000 \)[/tex].
[tex]\[ x = \frac{60 \pm \sqrt{3600 + 432000}}{6} \][/tex]
[tex]\[ x = \frac{60 \pm \sqrt{435600}}{6} \][/tex]
[tex]\[ x = \frac{60 \pm 660}{6} \][/tex]
Solving for the two possible values:
[tex]\[ x_1 = \frac{720}{6} = 120 \][/tex]
[tex]\[ x_2 = \frac{-600}{6} = -100 \][/tex]
Since the price of a cake cannot be negative, the valid solution is:
[tex]\[ x = 120 \text{ Tsh} \][/tex]
### (b) Modifying the Rectangular Garden:
Let's denote the length added to the shorter side and reduced from the longer side as [tex]\( y \)[/tex] meters.
1. Set up the problem:
- The original dimensions of the garden are 6 meters by 8 meters.
- Adjust the shorter side to [tex]\( 6 + y \)[/tex] meters.
- Adjust the longer side to [tex]\( 8 - y \)[/tex] meters.
- The new area of the garden is 45 square meters.
2. Form the equation:
[tex]\[ (6 + y)(8 - y) = 45 \][/tex]
3. Expand and simplify:
[tex]\[ 48 - 6y + 8y - y^2 = 45 \][/tex]
[tex]\[ - y^2 + 2y + 3 = 0 \][/tex]
Multiplying through by -1 to make it a standard quadratic equation:
[tex]\[ y^2 - 2y - 3 = 0 \][/tex]
4. Solve using the quadratic formula:
The quadratic formula is given by [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -3 \)[/tex].
[tex]\[ y = \frac{2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm 4}{2} \][/tex]
Solving for the two possible values:
[tex]\[ y_1 = \frac{6}{2} = 3 \][/tex]
[tex]\[ y_2 = \frac{-2}{2} = -1 \][/tex]
Here, [tex]\( y_1 = 3 \)[/tex] meters and [tex]\( y_2 = -1 \)[/tex] meters.
### Final Answers:
1. The price of the cake, [tex]\( x \)[/tex], is 120 Tsh.
2. The length to be added to the shorter side and reduced from the longer side to get an area of 45 square meters is [tex]\( y = 3 \)[/tex] meters or [tex]\( y = -1 \)[/tex] meters.
Let's denote the original price of the cake as [tex]\( x \)[/tex] Tsh.
1. Set up the problem:
- Amina has Tsh 1800 to spend.
- If the price decreases by Tsh 20, she can buy 3 more cakes.
- With the original price, the number of cakes she can buy is [tex]\( \frac{1800}{x} \)[/tex].
- With the decreased price [tex]\( x - 20 \)[/tex], the number of cakes she can buy is [tex]\( \frac{1800}{x - 20} \)[/tex].
2. Form the equation:
[tex]\[ \frac{1800}{x - 20} = \frac{1800}{x} + 3 \][/tex]
3. Eliminate the fractions:
Multiply through by [tex]\( x(x - 20) \)[/tex]:
[tex]\[ 1800x = 1800(x - 20) + 3x(x - 20) \][/tex]
4. Expand and simplify:
[tex]\[ 1800x = 1800x - 36000 + 3x^2 - 60x \][/tex]
[tex]\[ 0 = 3x^2 - 60x - 36000 \][/tex]
5. Form the quadratic equation:
[tex]\[ 3x^2 - 60x - 36000 = 0 \][/tex]
6. Solve using the quadratic formula:
The quadratic formula is given by [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -60 \)[/tex], and [tex]\( c = -36000 \)[/tex].
[tex]\[ x = \frac{60 \pm \sqrt{3600 + 432000}}{6} \][/tex]
[tex]\[ x = \frac{60 \pm \sqrt{435600}}{6} \][/tex]
[tex]\[ x = \frac{60 \pm 660}{6} \][/tex]
Solving for the two possible values:
[tex]\[ x_1 = \frac{720}{6} = 120 \][/tex]
[tex]\[ x_2 = \frac{-600}{6} = -100 \][/tex]
Since the price of a cake cannot be negative, the valid solution is:
[tex]\[ x = 120 \text{ Tsh} \][/tex]
### (b) Modifying the Rectangular Garden:
Let's denote the length added to the shorter side and reduced from the longer side as [tex]\( y \)[/tex] meters.
1. Set up the problem:
- The original dimensions of the garden are 6 meters by 8 meters.
- Adjust the shorter side to [tex]\( 6 + y \)[/tex] meters.
- Adjust the longer side to [tex]\( 8 - y \)[/tex] meters.
- The new area of the garden is 45 square meters.
2. Form the equation:
[tex]\[ (6 + y)(8 - y) = 45 \][/tex]
3. Expand and simplify:
[tex]\[ 48 - 6y + 8y - y^2 = 45 \][/tex]
[tex]\[ - y^2 + 2y + 3 = 0 \][/tex]
Multiplying through by -1 to make it a standard quadratic equation:
[tex]\[ y^2 - 2y - 3 = 0 \][/tex]
4. Solve using the quadratic formula:
The quadratic formula is given by [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -3 \)[/tex].
[tex]\[ y = \frac{2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm 4}{2} \][/tex]
Solving for the two possible values:
[tex]\[ y_1 = \frac{6}{2} = 3 \][/tex]
[tex]\[ y_2 = \frac{-2}{2} = -1 \][/tex]
Here, [tex]\( y_1 = 3 \)[/tex] meters and [tex]\( y_2 = -1 \)[/tex] meters.
### Final Answers:
1. The price of the cake, [tex]\( x \)[/tex], is 120 Tsh.
2. The length to be added to the shorter side and reduced from the longer side to get an area of 45 square meters is [tex]\( y = 3 \)[/tex] meters or [tex]\( y = -1 \)[/tex] meters.
