GCSE maths equation

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12345678

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A tennis court has an area of 224m^2. If the length were decreased by 1m and the width increased by 1m, the area would be increased by 1m^2. Find the dimensions of the court. Can somebody explain how you would work this out? much appreciated:). To start off with, I decided that x was the length and y was the width, so xy=224m^2. I then did (x-1)(y+1)=225m^2. so xy +x -y=226m^. What do I do next, or is what I have done wrong?
 
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12345678 said:
A tennis court has an area of 224m^2. If the length were decreased by 1m and the width increased by 1m, the area would be increased by 1m^2. Find the dimensions of the court. Can somebody explain how you would work this out? much appreciated:).
Click to expand...

Please see "Read before psoting".

You know that Area = Length times Width.

224 = LW for the tennis court.

If the dimensions are altered as stated, then the new length is (L - 1) and the new width is (W + 1). The new area would be (224 + 1).

Make a second equation with that information. Solve the system of equations.
 
12345678 said:
A tennis court has an area of 224m^2. If the length were decreased by 1m and the width increased by 1m, the area would be increased by 1m^2. Find the dimensions of the court. Can somebody explain how you would work this out? much appreciated:).
Click to expand...
We like to see your work, so we know where you are getting stuck!

You must know the formula for the area of a rectangle, A = L×W
The first rectangle is L×W = 224 m^2
The second rectangle is (L - 1m)×(W - 1m) = ...

That will give you two equations in two unknowns - can you solve that system?
 
wjm11 said:
Please see "Read before psoting".

You know that Area = Length times Width.

224 = LW for the tennis court.

If the dimensions are altered as stated, then the new length is (L - 1) and the new width is (W + 1). The new area would be (224 + 1).

Make a second equation with that information. Solve the system of equations.
Click to expand...
Do I expand the brackets in the second equation? So (L-1) x (w+1)= LW +L -W =226?
 
12345678 said:
A tennis court has an area of 224m^2. If the length were decreased by 1m and the width increased by 1m, the area would be increased by 1m^2. Find the dimensions of the court. Can somebody explain how you would work this out? much appreciated:). To start off with, I decided that x was the length and y was the width, so
xy = 224m^2..............(1)
I then did
(x-1)(y+1)=225m^2.
xy + x - y = 226m^2...(2)
What do I do next, or is what I have done wrong?
Click to expand...
Great so far.

The best way to solve the system of 2 equations is probably substitution.
Solve eq. (1) for y as a function of x,
then substitute that expression for y into eq. (2).
Because there will be an x in the denominator, you should multiply through by x, giving you a quadratic equation to solve for x.
Actually, the two solutions of the quadratic will be x and y.
 
DrPhil said:
Great so far.

The best way to solve the system of 2 equations is probably substitution.
Solve eq. (1) for y as a function of x,
then substitute that expression for y into eq. (2).
Because there will be an x in the denominator, you should multiply through by x, giving you a quadratic equation to solve for x.
Actually, the two solutions of the quadratic will be x and y.
Click to expand...
(1) y=224/x
(2) xy+x-y=226. If I substitute the expression in for y twice, I get x(244/x) +x-(224/x). This second part (+x-(224/x) does not seem right?. I understand that if I multiply x to get it off the bottom I get x^2 x224, but I don't understand the second part.
 
12345678 said:
(1) y=224/x
(2) xy+x-y=226. If I substitute the expression in for y twice, I get x(244/x) +x-(224/x). This second part (+x-(224/x) does not seem right?. I understand that if I multiply x to get it off the bottom I get x^2 x224, but I don't understand the second part.
Click to expand...
Yes, \(\displaystyle x\frac{224}{x}+ x- \frac{224}{x}= 224+ x- \frac{224}{x}= 226\). Mutiplying both sides by x gives \(\displaystyle 224x+ x^2- 224= 226x\) so you need to solve the quadratic equation \(\displaystyle x^2- 2x- 224= 0\).
 
12345678 said:
(1) y = 224/x
(2) xy + x - y = 226.
If I substitute the expression in for y twice, I get
(3) x(224/x) + x - (224/x) = 226 ... don't forget to include both sides of the equation.
This second part (+x-(224/x) does not seem right?. it is fine.
I understand that if I multiply x to get it off the bottom I get x^2 x224, but I don't understand the second part.
Click to expand...
Looking at eq. (3), the first term is just 224 m^2, which can be subtracted from both sides:

(3') x - (224/x) = 2

Now multiply all terms (both sides!) by x, and rearrange into the form of a quadratic equation.
 
DrPhil said:
Looking at eq. (3), the first term is just 224 m^2, which can be subtracted from both sides:

(3') x - (224/x) = 2

Now multiply all terms (both sides!) by x, and rearrange into the form of a quadratic equation.
Click to expand...
Oh, now I understand. using the quadratic formula, x=16. Therefore the dimensions are 16 and 14, as 244/16=14. Thanks for the help, much appreciated!
 
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