Can somebody show me how to work these problems....?

Question

Find three Consecutive odd integers such that the sum of all three is 42 less than the the product of the larger two........

Two Cars left an intersection at the same time. One traveled north. The other traveled 14 mi farther, but to the east. How far apart were they then, if the distance between them was 4 mi more than the distance traveled east? I have to use pytuagorean therom.

A lo has the shape of a right triangle with one leg 2 m longer than the other. The Hypotenuse is 2m less than twice the length of the shorter leg. Find the length of the shorter leg.

I'm not very good at word problems please help explain them...

Answer 1

a) we know they are consecutive odd integers, so we will use x, x+2, and x+4 to represent the 3 integers. We know that the sum of them is 42 less than the product of the larger 2, so...

x + x+2 + x+4 + 42 = (x+2) * (x+4)
3x + 48 = x^2 + 6x + 8
x^2 + 3x - 40 = 0
(x + 8)(x - 5) = 0
x = -8, 5

Since -8 is even, 5 is what we want... so our integers are 5, 7, 9

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We will let x represent the distance travelled north. The distance travelled east will then be x+14, and the distance between them is x+14+4 or x+18.

So now we have a right triangle with sides of length x and x+14, and a hypotenuse of x+18.

Using the pythagorean theorem, we know that:

x^2 + (x+14)^2 = (x+18)^2
x^2 + x^2 + 28x + 196 = x^2 + 36x + 324
x^2 - 8x - 128 = 0

now we use the quadratic formula to solve for x:

x = -b +/- (b^2 - 4ac)^.5 / 2a
x = 8 +/- (64 - 4(1)(-128))^.5 / 2
x = 8 +/- (576)^.5 / 2
x = 8 +/- 24 / 2
we know x is positive, so we will ignore the negative...
x = (8+24)/2 = 32/2 = 16 miles

But they are x + 18 miles apart, so 16+18 = 34 miles apart!
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Again we have a right triange. This time the sides are x and x+2. They hypotenuse is 2x-2.

So... x^2 + (x+2)^2 = (2x-2)^2

x^2 + x^2 + 4x + 4 = 4x^2 - 8x + 4
2x^2 - 4x = 0
2x (x-2) = 0

so 2x = 0 or x-2 = 0
x = 0, 2

we can't have a side of length 0, so our answer is 2m.

Hope this helps! :) Best of luck

Answer 2

The integers are x, y and z.
x + 2 = y, y + 2 = z (because of three consecutive odds. e.g. 1,3 and 5: 1 + 2 = 3, 3 + 2 = 5)

x + y + z = y*z - 42 (sum of all three is 42 less than the product of the larger two. sum of all three = x+y+z, product of the larger two = y*z, 42 less = -42)

Now you have a system of three equations:
x + 2 = y
y + 2 = z
x + y + z = y*z - 42
Substitute y by (x+2), because x + 2 = y:
x + (x+2) + z = (x+2)*z - 42
Substitute z by (y + 2) with y = x + 2:
x + (x+2) + ((x+2) + 2) = (x+2)*((x+2) + 2) - 42
Simplify and solve:
x + (x+2) + ((x+2) + 2) = (x+2)*((x+2) + 2) - 42
x + x + 2 + x + 2 + 2 = (x+2)*(x + 2 +2) - 42
3*x + 6 = (x+2)*(x+4) - 42
3*x + 6 = (x^2 + 4*x + 2*x + 8) - 42
3*x + 6 = x^2 + 6*x - 34
3*x = x^2 + 6*x - 40
0 = x^2 + 3*x - 40

Quadratic formula:
x^2 + px + q = 0
x = -p/2 +- Sqrt((p^2/4 - q)
x = -3/2 +- Sqrt(3^2/4 - (-40))
x = -1.5 +- Sqrt(9/4 + 40)
x = -1.5 +- 6.5
x = -8 and x = 5

-8 is not odd, so the solution has to be x = 5, then y = 5 + 2 = 7 and z = 7 + 2 = 9.
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If b is the distance travelled east, a the distance travelled north and c the distance between the cars:
b - 14 = a
b + 4 = c
a^2 + b^2 = c^2
(b-14)^2 + b^2 = (b+4)^2
Expand:
b^2 - 28*b + 14^2 + b^2 = b^2 + 8*b + 4^2 || -b^2
-28*b + 14^2 + b^2 = 8*b + 4^2 ||+28*b
14^2 + b^2 = 36*b + 4^2 ||-14^2
b^2 = 36*b - 180
0 = -b^2 + 36*b -180 ||*(-1)
0 = b^2 - 36*b + 180
Quadratic formula:
x = -p/2 +- Sqrt((p^2/4 - q)
b = 36/2 +- Sqrt(36^2/4 - 180)
b = 18 +- Sqrt(324 - 180)
b = 18 +- Sqrt(144)
b = 18 +- 12
b = 30 and b = 6

a = b - 14, a = 16 and a = -8 (can't be -8 because distance can't be negative, so b = 30)
c = b + 4, c = 34 and c = 10 (a^2 + b^2 = c^2, 16^2 + 30^2 = 34^2, so c = 34)
a = 16, b = 30, c = 34
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The shortest leg is x, the second one x+2 and the hypothenuse is 2*x-2:

x^2 + (x+2)^2 = (2*x-2)^2
x^2 + x^2 + 4*x + 4 = 4*x^2 - 8*x + 4 ||-4
x^2 + x^2 + 4*x = 4*x^2 - 8*x || -4*x
x^2 + x^2 = 4*x^2 - 12*x
2*x^2 = 4*x^2 - 12*x ||-2*x^2
0 = 2*x^2 - 12*x ||/2 (so you can use the quadratic formula)
0 = x^2 - 6*x + 0

x = 6/2 +- Sqrt(6^2/4)
x = 3 + Sqrt(9)
x = 6

Answer 3

1. Represent the consecutive integers as x, (x+2), and (x + 4). The sum is 3x + 6, and the product of the larger two is (x + 2)(x + 4), or x^2 + 6x + 8. Now add 42 to the sum to make it equal to the product:

3x + 48 = x^2 + 6x + 8, and solve the quadratic equation.

2. Represent the north miles as N. East miles are N + 14, and the hypotenuse (distance between them...draw this out and you'll see that it's a right triangle) is N + 14 + 4 (4 miles more than travelled east). Now use the pythagorean theorem to get the equation:

N^2 + (N+14)^2 = (N+18)^2 and solve

3. Once again, represent one of the legs as X (use the shorter one for example, and this will make it easier at the end, since this is the one you're looking for. In general, if you have a choice of which variable to make X, try and make it so X is your answer). The other leg is then X + 2, and the hypotenuse is 2X - 2 (twice the length of the short one, minus 2).

The equation:

X^2 + (X+2)^2 = (2X -2)^2

Answer 4

Let's say that x, y, and z are three consecutive odd integers. We're looking for the sum of all three to equal the product of the greater two minus 42. In other words:

xyz = yz - 42

Since they are consecutive odd integers, y = x+2 and z = y+2 = x+4. Plugging in these values gives us

x+(x+2)+(x+4) = (x+2)(x+4) - 42

Simplifying this gives

3x+6 = x^2 + 6x + 8 - 42

Get this equal to zero:

x^2 + 3x - 40 = 0

Factor this into

(x - 5)(x + 8)

Now the two possible answers for x are
x = 5 or x = -8

-8 is not an odd integer, so the answer that you're looking for is x=5. Plug this into y=x+2 and z=x+4 to find that our three answers are

5,7,9

Answer 5

They are fairly simple, you just have to substitute X for the unknown value in each one and turn it into an algebraic equation.

In the second you have cars. Try drawing it out that helps me. The missing figure is the distance the north car traveled. Make that X and create your equation.

The third is simpler since they say it is like a right angle triangle. Draw the triangle, insert the length for each side using X as the length of the shortest side and use Pythagoras Therom to create your equation. Then solve.

If you need more specific help, feel free to contact me.

Answer 6

1st question only

5, 7 n 9

Product of 2 large integer is 7x9=63
Sum of 3 integer is 21

63-21=42

5, 7 and 9 are consecutive integers