Systems of Equations in Two Variables

This is my study guide where I work problems on systems of equations in two variables.

Systems of Equations in Two Variables

A system of equations in two variables can have 0,1,2 or infinitely many solutions. A system of nonlinear equations can have any number of solutions. However, if the equations are all linear, then only 0,1 or infinitely many solutions are possible. 

Any system of linear equations in two variables can be written as:

\[a_1x + b_1y = c_1\]

And

\[a_2x + b_2y = c_2\]

The graph of this consists of two lines. Coincident lines are identical lines and indicate that the two equations are equivalent and have the same graph. The lines of this graph can be intersecting, coincident, or parallel.

A consistent system of linear equations has either one solution or infinitely many solutions. If the system has one solution, then the equations are independent, and they can be represented by lines that intersect at one point. If the system has infinitely many solutions, then the equations are dependent, and they can be represented by coincident lines. An inconsistent system has no solutions and can be represented by parallel lines. 

The graphs of the two equations are distinct lines that intersect at one point. The system is consistent and there is one solution. It is given by the coordinates of the point of intersection. The equations are independent.

The graphs of the two equations are the same line. The system is consistent. There are infinitely many solutions and the equations are dependent. 

The graphs of the two equations are distinct parallel lines. The system is inconsistent and there are no solutions. 

Example 1

Use elimination to solve the system of equations. Identify the system as consistent or inconsistent. Then, state whether the equations are dependent or independent. 

\[2x - y = -4\]

\[3x + y = -1\]

We eliminate the ‘y’ variable by adding the equations.

This leaves us with:

\[5x = -5\]

\[x = -1\]

The ‘y’ variable can be found by substituting ‘x=-1’ in either of the original equations.

\[2x - y = -4\]

\[2(-1) - y = -4\]

\[-2 - y = -4\]

\[- y = -2\]

\[y = 2\]

The solution is -1,2

There is a unique solution so the system is consistent and the equations are independent.

Example 2

Solve the system of equations.

\[4x - y = 10\]

\[-4x + y = -10\]

Add the equations

\[0 = 0\]

The equations are equivalent.

There are infinitely many solutions.

The system is consistent and the equations are dependent.

Example 3

Solve the system of equations.

\[x - y = 6\]

\[x - y = 3\]

Add the equations together.

\[0 = 3\]

The equation 0=3 is never true.

So, there are no solutions and the system is inconsistent. 

Example 4

Solve the system of equations.

\[\frac{x+y}{2} = 6050\]

\[x - y = 3900\]

Multiply the first equation by 2 then add both equations together.

\[x + y = 12100\]

\[x - y = 3900\]

\[2x = 16000\]

\[x = 8000\]

Now, we substitute ‘x=8000’ into one of the original equations.

\[8000 - y = 3900\]

\[y = 4100\]

We have a solution of ‘8000,4100’

Example 5

Solve the system of equations.

\[2x - 3y = 18\]

\[5x + 2y = 7\]

Our goal is to eliminate the ‘y’ variable.

So, multiple the first equation by 2 and the second equation by 3.

\[4x - 6y = 36\]

\[15x + 6y = 21\]

Now, we add the equations together.

\[19x = 57\]

\[x = 3\]

Substitute ‘x=3’ into one of the first equations.

\[4(3) - 6y = 36\]

\[12 - 6y = 36\]

\[-6y = 24\]

\[y = -4\]

Our solutions are ‘3,-4’

Example 6

Solve the system of equations.

\[5x + 10y = 10\]

\[x + 2y = 2\]

We need our ‘y’ terms to be the same.

So, multiply the second equation by 5

\[5x + 10y = 10\]

\[5x + 10y = 10\]

\[0 = 0\]

This is always true so the equations are dependent and there are infinitely many solutions. 

Example 7

Solve the system of equations of a nonlinear system.

\[x^2 + y^2 = 4\]

\[2x^2 - y = 7\]

If we multiply each side of the first equation by and subtract the second equation, we can eliminate the ‘x’ variable.

\[2x^2 + 2y^2 = 8\]

\[2x^2 - y = 7\]

\[2y^2 + y = 1 \]

Now, we solve for ‘y’.

\[2y^2 + y - 1 = 0\]

Factor

\[(y+1)(2y-1)=0\]

\[y=-1 \text{or} y = \frac{1}{2}\]

We can now substitute the values back in and it looks like we will get some weird results.