Answer :
1) the cost of 1 kg of crabs is [tex]\( \frac{140}{x} \)[/tex].
2) The cost of 1 kg of fish is [tex]\( \frac{140}{x + 3} \)[/tex]
3) Quadratic equation: [tex]\(3x^2 + 9x - 84 = 0\)[/tex]
4) Solutions to the equation: x = 4 (the other solution is not applicable)
5) Quantity of fish bought: 7 kg and Quantity of crab bought: 4 kg
1) The cost of x kg of crabs is $140.
So, the cost of 1 kg of crabs is [tex]\( \frac{140}{x} \)[/tex].
2) She received 3 kg more fish than crabs.
If she bought x kg of crabs, then she bought (x + 3) kg of fish with the same $140.
So, the cost of 1 kg of fish is [tex]\( \frac{140}{x + 3} \)[/tex].
3) The cost of 1 kg of fish is $15 less than the cost of 1 kg of crab.
This gives us the equation:
[tex]\[ \frac{140}{x + 3} = \frac{140}{x} - 15 \][/tex]
To solve this equation, let's first get rid of the fractions:
[tex]\[ 140x = 140(x + 3) - 15x(x + 3) \][/tex]
[tex]\[ 140x = 140x + 420 - 15x^2 - 45x \][/tex]
[tex]\[ 0 = -15x^2 - 45x + 420 \][/tex]
Multiplying through by -1 to make the leading coefficient positive:
[tex]\[ 0 = 15x^2 + 45x - 420 \][/tex]
Dividing the entire equation by 15 to simplify:
[tex]\[ 0 = x^2 + 3x - 28 \][/tex]
This reduces to the quadratic equation [tex]\(3x^2 + 9x - 84 = 0\)[/tex].
4) Now, let's solve [tex]\(3x^2 + 9x - 84 = 0\)[/tex].
Using the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \][/tex]
where a = 3 , b = 9 , and c = -84.
Plugging in the values:
[tex]\[ x = \frac{{-9 \pm \sqrt{{9^2 - 4 \cdot 3 \cdot (-84)}}}}{{2 \cdot 3}} \][/tex]
[tex]\[ x = \frac{{-9 \pm \sqrt{{81 + 1008}}}}{{6}} \][/tex]
[tex]\[ x = \frac{{-9 \pm \sqrt{{1089}}}}{{6}} \][/tex]
[tex]\[ x = \frac{{-9 \pm 33}}{{6}} \][/tex]
So, we have two possible solutions:
[tex]\[ x_1 = \frac{{-9 + 33}}{{6}} = \frac{24}{6} = 4 \][/tex]
[tex]\[ x_2 = \frac{{-9 - 33}}{{6}} = \frac{-42}{6} = -7 \][/tex]
Since we're dealing with quantities of items, the number of kilograms cannot be negative.
Therefore, we take x = 4 as our solution.
5) Substituting x = 4 into the expression for the cost of 1 kg of crabs,
we find that each kilogram costs [tex]\( \frac{140}{4} = $35 \)[/tex].
Substituting x = 4 into the expression for the cost of 1 kg of fish,
we find that each kilogram costs [tex]\( \frac{140}{4 + 3} = \frac{140}{7} = $20 \)[/tex].
So, she bought 4 kilograms of crabs and 7 kilograms of fish.
Answer: 13kg
Step-by-step explanation: she bought a total of 14/3 + 25/3 = 39/3 = 13 kg of fish and crabs.
