Simultaneous Equations

Question

Solve the following system of simultaneous equations: 

 

(x-3)^2+(y-3)^2=50

x=y

 

And another question: 

 

A group of students is learning to use a compass by following a set of directions to arrive at an unknown
location. The directions are as follows:
Start at the flagpole.
1. Go 20 meters north, then 10 meters west.
2. Go 10 meters south, then 50 meters east.
3. Go 10 meters south, then 10 meters east.
4. Go 10 meters south, then 10 meters west.

 

Where did they end up? ------ meters ----, and ----- meters -------. (dashed lines are blanks)

How far are they from where they started? ------- meters

 

 

Please explain how to do it step by step. Thanks!

 

 

Answer 1

Question 1:

Since x=y we can substitute a x for y in the first equation giving us

(y-3)^2 + (y-3)^2 = 50

leading to: 2(y-3)^2 = 50

dividing both sides by 2 leaves us with

(y-3)^2 = 25.

Two ways to proceed from here:

 

option 1:multiply out (y-3)^2 and set equal to 0 and then use quadratic equation to get solutions.

 

option 2: take the square root of both sides of (y-3)^2 = 25, giving the next line of

(y-3) ± 5. Now add 3 to both sides to solve for y.

y = 3 + 5 and y = 3 - 5

giving answers of y = -2 and y = 8. Since x = y, x= -2 and 8 also.

 

You would have to use option 1 if the equation didn't equal a nice square root like 25.

 

Using option 1:

we must change (y-3)^2 = 25 to (y-3)(y-3)=25

multiplying this out leads to y^2 - 6y + 9 = 25. Set this equal to zero by subtracting 25 from both sides

y^2 - 6y - 16 = 0. This is factorable. If you can see this is equal to (y + 2)(y - 8) = 0, then you can see that y = -2 or 8.

If it is too difficult to figure out the factors you can always use the quadratic formula:

[-b ± √(b^2 -4ac)]/2a

with ay^2 + by + c = 0

Therefore using y^2 - 6y - 16 = 0; The values are a=1, b=-6, c=-16

Plug those values into [-b ± √(b^2 -4ac)]/2a and you will get -2 and 8 as answers.

 

Question 2

 

If you carefully plot out the directions step-by-step on graph paper or using a ruler you will end up 10m South and 40m East. From your origin go down one increment and make a mark, then from there move east(right) 4 increments (40m) and make a mark. Join up the points to create a right triangle. The length of the hypotenuse is how far there are from where they started. To solve for that use the Pythagorean Theorem a^2+b^2=c^2 where c is the hypotenuse.

So 10^2+40^2=c^2 solve for c.

 

 

Answer 2

the numbers are all in meters

 

20 up, 10 left, 10 down, 50 right (this takes you through the middle of the first "20 up" and 40 to the right of the first "20 up"), 10 down brings you 40 away from where you started on the horizontal, 10 right now brings you 50 away from where you started on the horizontal, 10 down and then 10 left brings you 40 m east and 10 m south of your starting point

 

you have a right triangle with legs of 40 m and 10 m

now find the hypotenuse and you will know how far they are from where they started

x=hypotenuse

x²=40²+10²

x²=1600+100

x²=1700

x=√1700

x=41.23105 feet from where they started

 

(x-3)^2+(y-3)^2=50

x=y

substitute

(x-3)^2+(x-3)^2=50

2(x-3)^2=50

(x-3)^2=25

x^2-6x+9=25

x^2-6x+9-25=0

x^2-6x-16=0

factor

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

x-8=0

x=8 and y=8 because x=y

x+2=0

x=-2 and y=-2 because x=y