Construct a system of two linear equations

Question

Construct a system of two linear equations where (-5,4) is a solution to the first equation but not to the second equation, and where  (5,-2) is the solution to your system. Step 1 plot the points given to you and write the equation of that line. Call it L1.Step 2 on the graph draw a second line that will create a system of equations with the desired solution. Step 3 write the equation of your new line. Call it L2. Follow up questions 1 explain how your graph shows that your system satisfies the required conditions. 2 should you have the same equation as your partner for L1? Explain why or why not. 3 should you have the same equation as your partner for L2? Explain why or why not 

Answer 1

Step 1:

 

Plot the points (-5, 4) and (5, -2) on the same coordinate system.  (5, -2) is the point of intersect of the two lines.

 

If we use the point-slope form of a line using the two given points, we can find the line L1.  This is our reference line.   You also use the y-intercept and slope of the line to get your equation.

 

L1 = mx - mx₁ + y₁

 

where:

m = slope

(x₁ , y₁) is any of the given points

 

 

First, find the slope between the points.

 

m = (-2 - 4) / (5 - (-5))

m = -6 / 10

m = -3 / 5

 

Then using any of the points perhaps (5, -2), we get the line

 

L1 = (-3/5)x - (-3/5)(5) + (-2)

L1 = (-3/5)x + 1

 

This is the first equation.

 

 

Step 2:

 

Draw any line (with a slope not equal to -3/5) that intersects L1 at (5, -2).

 

 

Step 3:

 

Use the y-intercept and slope from the last step to write the equation of the line in slope-intercept form.  Name this line L2.

 

 

The next parts can be completed on your own.  Just use the solutions above.

Answer 2

There are multiple solutions to this problem, but the way you approach it is pretty consistent.  

1). Line L1 will pass through points (-5, 4) and (5, -2).  Once you plot the points, draw a straight line through them.

 

This is your first line and will satisfy this problem as long as line number two is not identical to it.

 

2). Find the equation of the line.  This can be done by finding the slope of the line and the y intercept.  Recall that the equation for the slope of a line is the change in y (y2 - y1), or in this case -2 - 4, divided by the change in x (x2 - x1), or 5 - (-5).  

 

You can find the y intecept using y = mx + b where m is the slope you just found.  

 

3). Construct your second equation.  This can literally be any line that passes through the point (5, -2).  For my solution I simply used y = -2, but your teacher might not appreciate this solution. A very simple strategy to find a decent line is to simply plug the point (5, -2) into x + y = ?.  In this case it comes to x + y = 3. Then manipulate it into -x + 3 = y. Easy peezy.

 

Boom.  You got yourself a system of equations. 

 

 

Answer 3

We know for one line it contains

(-5,4) and (5,-2)

 

The equation of that line can be found by

using the point slope formula y-y₁ = m(x-x₁)

First find the slope.

m = (-2-4)/(5-(-5)) = -6/10 = -3/5

 

Choose one of the points as (x₁,y₁) and plug in.

Let's use (-5,4)

y-4 = (-3/5)(x-(-5))

y-4 = (-3/5)(x+5)

5(y-4) = -3(x+5)

5y-20 = -3x - 15

3x + 5y = 5

 

We now have the equation of the 1st line.

This line includes (-5, 4) and (5, -2)

 

Now let's find the equation of a line that

includes (5, -2) but not (-5,4).

This will be a line with different slope that

includes (5, -2)

 

y = mx + b

 

Pick a slope that is not -3/5

For ease let m = 1

 

y = x + b

Plug in the point (5,-2) for (x,y) and solve for b

-2 = 5 + b

b = -7

y = x-7

 

We now have the equations for 2 lines that

intersect at (5, -2)

 

1)  3x + 5y = 5

2)  y = x-7

 

Note: by substitution we can verify the solution

3x + 5(x-7) = 5

3x + 5x - 35 = 5

8x = 40

x = 5

y = x-7 = -2

solution (5, -2)

 

Since we could have chosen any slope other than

-3/5 to create the 2nd equation then it is entirely

possible for separate students to have a different

set of equations for their system.