How to find the solution for a system of equations - SAT Math

After 20 minutes the first train would have traveled 20 miles. Let x be the amount of time elapsed. When 20 + 60x = 100x you will have the time in hours. 20 = 40x, x = 0.5 hrs. 0.5 hrs = 30 minutes.

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.

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

7_x_ + y = 25 – 6_x_ + y = 23: 7_x_ – 6_x_ = 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:

6_x_ + y = 23 → y = 23 – 6_x_.

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

Subtract and from the both sides to get .

Divide both sides by , to get

You can solve this problem in a number of ways, but one way to solve it is by using substitution. You can begin to do that by solving for in the first equation:

Now, you can substitute in that value of into the second equation and solve for :

Let's consider this equation as adding a negative 3 rather than subtracting a 3 to make distributing easier:

Distribute the negative 3:

We can now combine like variables and solve for :

We are essentially presented with a system of equations. To solve for y, we will need to solve the system. The easiest way to solve this particular system is by adding the equations together.

First, multiply the second equation by 2.

Adding the two equations together will allow you to cancel the x values and solve for y.

If y equals 2, then 4y will be equal to 8.

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.

Substitution

Solve the second equation for x:

Substitute this expression for x in the first equation:

Solve for y:

Substitute this value for y in any equation and solve for x.

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

To solve this problem, we translate the given information into two equations and then solve both simultaneously. If we let F represent the number of tanks that hold four fish and S represent the number of tanks that hold six fish, the problem tells us that F+S=15. The problem also tells us that 4F (the total number of fish in the 4-fish tanks) plus 6S (the total number of fish in the six-fish tanks) equals 70 (the total number of fish in the aquarium).

Thus we have the following system of equations:

F+S=15

4F+6S=70

Multiplying the first equation by -4 and combing it with the second gives 2S = 10, as seen below:

\[-4F-4S=-60 (the first equation times -4)\]

+ \[4F+6S=70 (the second equation)\]

2S = 10

Therefore, S, the number of tanks that hold 6 fish, is 5.

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

To solve for y, first eliminate x by adding the two equations together such that the x’s factor out:

4x + 9y + 7 = 0

(–2)2x – (–2) 3y + (–2) 6 = (–2)0 (Multiply this equation by a factor of –2 so that 4x – 4x = 0)

Therefore, the two equations added together are:

(4x + 9y + 7 = 0) + (–4x –(–6)y +(–12) = 0) = (0 + 15y – 5 = 0)

15y = 5

y = 1/3

Solve for by merging the equations so that gets factored out. To do so, multiply the lower equation by (so that at the top is subtracted by )

First, solve this equation for y and then substitute the answer into the second equation:

Now substitute into the second equation and solve for x:

To solve for x, add the coefficients on the x variables together then divide both sides by three.

Set as the number of people and as the number of dogs.

The head equation is going to be:

The legs equation is going to be:

Manipulating the first equation to solve for results in the following.

Plugging in

into the second equation and solving for , you get

Setting x and the number of quarters he has and y as the numbver of nickels. x + y = 14 (total coins), 0.25x + 0.05y = 1.50 (total amount). Substituting x = 14 – y from the first equation into the second, we get y = 10. Therefore Joey has 10 nickels.

Setting t as the time elapsed we need to find when 8t = 12 + 4t (this is the distance traveled by the ball compared to the distance traveled by the player+difference from origin). Solving for t we get a travel time of 3 seconds. If the player runs for 3 seconds at 4m/s, the player travels 12m before receiving the ball.