System of equations solution methods, examples, exercises

The ecuation systems They consist of two or more equations with several variables that must have a common solution. They are frequent, because in practice there are numerous situations that depend on many factors, which are related in various ways.

In general, a system of equations has the following form, where each function represents one of the conditions that the solution must satisfy:

Let's see an example: suppose that you need to make rectangular sheets of paper whose area is 180 cm^two and that have a perimeter of 54 cm. What should be the dimensions of the sheet?

To answer the question we take into account that the dimensions of a rectangular sheet are two: width and height. This means that we have 2 variables to which we will give the usual names of x and Y.

And these variables must satisfy the two conditions imposed at the same time:

-First condition: the area of ​​the sheet is 180 cm^two. This will be the first function: F₁.

-Second condition: the perimeter or contour of the sheet must be 54 cm. This is the second function F_two.

For each condition, an equation is established using algebraic language. The area A of a rectangular sheet is obtained by multiplying width by height:

A = x.y = 180 cm^two

And the perimeter P results from adding the sides. Since the perimeter is the sum of the sides:

P = 2x + 2y = 54 cm

The resulting system of two equations and two unknowns is:

xy = 180

2 (x + y) = 54

We need two numbers whose product is 180 and the double product of their sum is 54, or what is the same: added together they have to give 27. These numbers are 12 and 15.

In the section of solved exercises we will offer the detailed method to find these values, meanwhile the reader can easily verify by substituting, that they effectively satisfy both equations.

Article index

• 1 Examples of applications of systems of equations
• 2 Methods of solving systems of equations
  • 2.1 Substitution method
  • 2.2 Method of reduction or elimination
  • 2.3 Equalization method
  • 2.4 Graphical method
• 3 Exercises
  • 3.1 - Exercise resolved 1
  • 3.2 - Exercise resolved 2
• 4 References

Examples of applications of systems of equations

The situation proposed above contains 2 variables, and at least 2 equations are required to find them. There are systems with many more variables, but in any case, if the system has n of them, it requires at least n equations independent of each other (one cannot be a linear combination of the others) to find the solution, if it exists.

As for the applications, they are numerous. Here are some in which systems of equations prove their usefulness:

-Finding the currents that flow through a circuit using Kirchoff's laws.

-In land and air transport to establish departure and arrival times.

-Find the magnitudes of forces in dynamic or static systems subject to multiple interactions.

-To know the amount of items sold during a certain period of time, or in factories, to determine the dimensions of objects so that they satisfy certain conditions in terms of surface or volume.

-When determining how to distribute a capital in various investments.

-Establish the rates for various services, for example telecommunications or shows and know the amount of money collected (see resolved example 2)

Methods of solving systems of equations

Method replacement

-An equation is chosen and one of the variables is solved.

-Then we have to substitute the cleared variable in another equation. Then this variable disappears from there and if the system has two equations and two unknowns, an equation is left with a variable that can already be solved.

-If the system has more than two variables, we must solve for a third unknown from another equation and replace it as well.

An example of the application of this method is in solved exercise 1.

Reduction or elimination method

This method consists of adding or subtracting equations to eliminate one or more variables and leave only one. To do this, it is convenient to multiply the equations by a factor such that when adding with another equation, the unknown disappears. Let's see an example:

3x^two - Y^two = 11

x^two + 4y^two = 8

We multiply the first equation by 4:

12x^two - 4y^two = 44

x^two + 4y^two = 8

When adding them, the unknown disappears Y, remaining:

13x^two = 52

x^two = 4

Therefore x₁ = 2 and x_two = -2. With these values ​​the reader can check that and₁ = 1 and y_two = -1

Equalization method

When the system is two equations with two unknowns:

-Choose an unknown and solve for both equations.

-The results are equalized, which allows obtaining a single equation with a single unknown.

-This equation is solved and the result is substituted in one of the previous clearances to obtain the value of the other unknown..

This method will be applied in solved exercise 2 of the following section.

Graphical method

This method consists of graphing the curves that each equation represents. The point of intersection is the solution of the system. The following example shows the graphical solution of the system:

x^two + Y ^two = 1

2x + 4y = 0

The first of the equations is a circle of radius 1 centered at the origin and the second is a line.

The intersection of both are the two points shown in blue. The reader can verify that by substituting the coordinates of the points in the equations above, an equality is obtained.

Training

- Solved exercise 1

You need to make rectangular sheets of paper with an area of ​​180 cm^two and with a perimeter of 54 cm. What should be the dimensions of the sheet?

Solution

The system to solve is:

xy = 180

2 (x + y) = 54

The second equation can be simplified to x + y = 27, therefore:

xy = 180

x + y = 27

Solve for one of the unknowns in the second equation:

y = 27 - x

Clearance is replaced in the first:

(27 -x) = 180

Applying the distributive property:

-x^two + 27x = 180

Multiplying by (-1) on both sides of the equation and sending the 180 to the left side:

x^two - 27x +180 = 0

The result is an equation of the second degree in x, which is solved by the formula:

With a = 1, b = -27 and c = 180

- Exercise solved 2

An amusement park has the following entrance fees: children $ 1.5 and adults $ 4. In one day there were 2,200 visitors, raising $ 5050. Find the number of children and adults who visited the park that day.

Solution

Be x the number of children and Y the number of adults. We can establish the first of the equations knowing that the sum of both must be 2200:

x + y = 2200.

Now we go with the money raised. The price of the children's ticket is $ 1.5 for each child, by multiplying this value by x, the number of children, we will have the amount for child ticket:

1.5x = money raised for children's tickets

And if we multiply $ 4 per adult by the number and number of adult visitors, we get the total money for all adults:

4y = money raised by adult tickets

We add this together to get $ 5050:

1.5x + 4y = 5050

Our system of equations is:

x + y = 2200

1.5x + 4y = 5050

Let's solve it by equalization. We isolate the variable y from the first and the second equation:

y = 2200 - x

y = (5050 - 1.5 x) / 4

We match both expressions:

2200 - x = (5050 - 1.5x) / 4

We multiply everything by 4 to eliminate the fraction:

8800 - 4x = 5050 - 1.5x

We group the terms with x on the left and the pure numbers on the right:

-4x + 1.5x = 5050 - 8800

-2.5x = -3750

x = 1500 children.

We substitute this value in y = 2200 - x to find the number of adults:

y = 2200 - 1500 = 700 adults.

References

1. CK-12. Systems of equations and inequalities. Recovered from: ck12.org.
2. Hoffman, J. Selection of Mathematics Topics. Volume 2.
3. Jiménez, R. 2008. Algebra. Prentice hall.
4. Stewart, J. 2006. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning.
5. Zill, D. 1984. Algebra and Trigonometry. Mcgraw hill.