We can factor out a , leaving .

From there we can factor again to 

.

We need to solve this equation by factoring. 

 or 

Now, plug these values back in individually to make sure they check.

For x=-5: 

For x=4:

Both answers check. 

Since the left hand side of the problem is not set equal to 0, we cannot simply set each term equal to 0 and solve. Instead, we need to multiply out the left side, subtract the -2 over to the right side, and then re-factor. 

To double check our answers, plug in -4 and -3 into the original problem. 

For x=-4:

For x=-3:

Both answers check. 

We need two numbers that add to  and multiply to be . Trial and error will show that  and  add to  and multiply to .

The expression  can be factored. 

We must find two numbers that added together equal -7, and multiplied together equal 60. Those two numbers are -12 and 5. 

Now we can set both terms equal to zero and solve for a.

, 

This expression can be factored by grouping. 

x² can be factored out of the first two terms. 

-4 can be factored out of the second two terms. 

Now we can factor out the term (x-3). 

x² - 4 is a difference of squares. It can be factored into (x + 2)(x - 2). 

You can subtract the second equation from the first equation to eliminate y:

7x + y = 25 – 6x + y = 23: 7x – 6x = x; y – y = 0; 25 – 23 = 2

x = 2

You could also solve one equation for y and substitute that value in for y in the other equation:

6x + y = 23 → y = 23 – 6x.

7x + y = 25 → 7x + (23 – 6x) = 25 → x + 23 = 25 → x = 2

We can add these equations to one another.

(7x + 3y = 20) + (–4x – 6y = 11) = (3x – 3y = 31)

If the point of intersection lies on all three lines, then we should be able to select any two lines, find their point of intersection, and come up with the same point of intersection each time. In other words, the point of intersection of the first two lines must be the point of intersection of the second and third lines. 

Let's consider the first and second lines. We can solve the system of equations by substituting the value of y from the second equation into the first.

y = 2x + 3

x + 2(2x + 3) = 1

x + 4x + 6 = 1

5x = -5

x = -1

y = 2(-1) + 3 = 1

The point of intersection of the first two lines is (-1,1).

Now we can find the point of intersection of the second and third lines. Again, we can substitute the value of y from the second equation into the third.

y = 2x + 3

4x - 3(2x + 3) = 2

4x -6x - 9 = 2

-2x = 11

x = -11/2

y = 2(-11/2)+3 = -8

Thus, the second and third lines intersect at (-11/2,-8).

Because the point of intersection between the first and second line does not coincide with the point of intersection between the second and third, there is no point that is common to ALL three lines. Thus, there is no point of intersection.

Set x as the number of sheep and y as the number of chicken. This gives us x+y=50 and 4x+2y=140. We want to solve for x. Solving the first equation we get y=50-x. Substitute that into the second you have 4x+2(50-x)=140. Multiplying it out gives 4x+100-2x=140. So 2x+100=140. 2x=40, x=20. Giving 20 sheep.