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A. Amina has Tsh. 1800 to spend on cakes. If the price of a cake is decreased by Tsh. 20, she can buy 3 more cakes. Find the price of the cake.

B. A rectangular garden is 6 meters wide and 8 meters long. What length should be added to the shorter side and reduced from the longer side of the garden to have an area of 45 square meters? (Use the quadratic formula).

Sagot :

### (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.
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