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.

The first thing to do is to write an algebrai equation for the problem:

8x – 9 = 5 – 10

8x – 9 = –5

8x = 4  

x = 1/2

Thus, 4 * x = 2

Add the two equations:

4x - 5y = 12 plus

6y - 3x = -6:

4x - 5y + (6y - 3x) = 12 + (-6)

4x - 3x + 6y - 5y = 12 - 6

x + y = 6

Find out how much a volunteer can produce in 3 hours.

6 tables/hour * 3 hours = 18 tables/hour

180 table need to be setup.  If one volunteer can setup 18 in 3 hours, then 10 volunteers will take care of the 180 tables.

2 auction galleries/hour * 3 hours = 6 galleries/hour

2 volunteers will be able to complete 12 auction galleries

10 + 2 = 12 volunteers

Solve for x, then plug into the formula to find the value.  x = 28 – 12 = 16

(3 * 16 + 2) * (–16 +10) = –300