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Solve Systems of Linear Equations with Two Variables

Determine whether an ordered pair is a solution of a system of equations

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations.

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

A linear equation in two variables, such as \(2x+y=7,\) has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs \((x,y)\) that make both equations true. These are called the solutions of a system of equations.

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example

Try it.

Determine whether the ordered pair is a solution to the system \(\{\begin{array}{l}x-y=-1 \\ 2x-y=-5\end{array}.\)

ⓐ \((-2,-1)\) ⓑ \((-4,-3)\)

Solution

ⓐ

ⓑ

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

How to Solve a System of Equations by Graphing

Try it.

Solve the system by graphing \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)

Solution

The steps to use to solve a system of linear equations by graphing are shown here.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example

Try it.

Solve the system by graphing: \(\{\begin{array}{l}3x+y=-1 \\ 2x+y=0\end{array}.\)

Solution

We’ll solve both of these equations for \(y\) so that we can easily graph them using their slopes and y-intercepts.


Solve the first equation for y.
Find the slope and y-intercept.
Solve the second equation for y.
Find the slope and y-intercept.
Graph the lines.
Determine the point of intersection.	The lines intersect at \((-1,2).\)
Check the solution in both equations.
	The solution is \((-1,2).\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

How to Solve a System of Equations by Substitution

Try it.

Solve the system by substitution: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)

Solution

Be very careful with the signs in the next example.

Example

Try it.

Solve the system by substitution: \(\{\begin{array}{l}4x+2y=4 \\ 6x-y=8\end{array}.\)

Solution

We need to solve one equation for one variable. We will solve the first equation for y.


Solve the first equation for y. Substitute \(-2x+2\) for y in the second equation.
Replace the y with \(-2x+2.\)
Solve the equation for x.
Substitute \(x=\frac{5}{4}\) into \(4x+2y=4\) to find y.
	The ordered pair is \((\frac{5}{4},-\frac{1}{2}).\)
Check the ordered pair in both equations.
	The solution is \((\frac{5}{4},-\frac{1}{2}).\)

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d.

\[\begin{array}{llllll}\text{if} & & & a & = & b \\ \text{and} & & & c & = & d \\ \text{then} & & & a+c & = & b+d.\end{array}\]

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\[\begin{array}{l}\{\underset{\text{—————}}{\begin{array}{l}3x+y=5 \\ 2x-y=0\end{array}} \\ \\ 5x\ =5\end{array}\]

The y’s add to zero and we have one equation with one variable.

\[\{\begin{array}{l}\ x+4y=2 \\ 2x+5y=-2\end{array}\]

How to Solve a System of Equations by Elimination

Try it.

Solve the system by elimination: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)

Solution

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

\[\begin{array}{lllllllllllllllllllllllllllllll}\text{Choose the Most Convenient Method to Solve a System of Linear Equations} \\ \begin{array}{lllllll}\underset{\text{————}}{\text{Graphing}} & & & \underset{\text{—————}}{\text{Substitution}} & & & \underset{\text{—————}}{\text{Elimination}} \\ \text{Use when you need a} & & & \text{Use when one equation is} & & & \text{Use when the equations are} \\ \text{picture of the situation.} & & & \text{already solved or can be} & & & \text{in standard form.} \\ & & & \text{easily solved for one} & & & \\ & & & \text{variable.} & & & \end{array}\end{array}\]

Example

Try it.

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ \(\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\) ⓑ \(\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\)

Solution

ⓐ

\[\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\]

Since both equations are in standard form, using elimination will be most convenient.

ⓑ

\[\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\]

Since one equation is already solved for y, using substitution will be most convenient.

Key Concepts

How to solve a system of linear equations by graphing.
1. Graph the first equation.
2. Graph the second equation on the same rectangular coordinate system.
3. Determine whether the lines intersect, are parallel, or are the same line.
4. Identify the solution to the system.
  If the lines intersect, identify the point of intersection. This is the solution to the system.
  If the lines are parallel, the system has no solution.
  If the lines are the same, the system has an infinite number of solutions.
5. Check the solution in both equations.
How to solve a system of equations by substitution.
1. Solve one of the equations for either variable.
2. Substitute the expression from Step 1 into the other equation.
3. Solve the resulting equation.
4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
5. Write the solution as an ordered pair.
6. Check that the ordered pair is a solution to both original equations.
How to solve a system of equations by elimination.
1. Write both equations in standard form. If any coefficients are fractions, clear them.
2. Make the coefficients of one variable opposites.
  Decide which variable you will eliminate.
  Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
6. Write the solution as an ordered pair.
7. Check that the ordered pair is a solution to both original equations.
  \(\begin{array}{lllllllllllll} \\ \\ \text{Choose the Most Convenient Method to Solve a System of Linear Equations} \\ \begin{array}{lllllll}\underset{\text{—————}}{\text{Graphing}} & & & \underset{\text{———————}}{\text{Substitution}} & & & \underset{\text{———————}}{\text{Elimination}} \\ \begin{array}{l}\text{Use when you need a} \\ \text{picture of the situation.}\end{array} & & & \begin{array}{l}\text{Use when one equation is} \\ \text{already solved or can be} \\ \text{easily solved for one} \\ \text{variable.}\end{array} & & & \begin{array}{l}\text{Use when the equations are} \\ \text{in standard form.}\end{array}\end{array}\end{array}\)

Solve Systems of Linear Equations with Two Variables

Determine Whether an Ordered Pair is a Solution of a System of Equations

In the following exercises, determine if the following points are solutions to the given system of equations.

Try it.

\(\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\)

ⓐ \((3,1)\) ⓑ \((-3,4)\)

Solution

ⓐ yes ⓑ no

Try it.

\(\{\begin{array}{l}-3x+y=8 \\ -x+2y=-9\end{array}\)

ⓐ \((-5,-7)\) ⓑ \((-5,7)\)

Try it.

\(\{\begin{array}{l}x+y=2 \\ y=\frac{3}{4}x\end{array}\)

ⓐ \((\frac{8}{7},\frac{6}{7})\) ⓑ \((1,\frac{3}{4})\)

Solution

ⓐ yes ⓑ no

Try it.

\(\{\begin{array}{l}2x+3y=6 \\ y=\frac{2}{3}x+2\end{array}\)
ⓐ \((-6,2)\) ⓑ \((-3,4)\)

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

Try it.

\(\{\begin{array}{l}3x+y=-3 \\ 2x+3y=5\end{array}\)

Solution

\((-2,3)\)

Try it.

\(\{\begin{array}{l}-x+y=2 \\ 2x+y=-4\end{array}\)

Try it.

\(\{\begin{array}{l}y=x+2 \\ y=-2x+2\end{array}\)

Solution

\((0,2)\)

Try it.

\(\{\begin{array}{l}y=x-2 \\ y=-3x+2\end{array}\)

Try it.

\(\{\begin{array}{l}y=\ \ \frac{3}{2}x+1 \\ y=-\frac{1}{2}x+5\end{array}\)

Solution

\((2,4)\)

Try it.

\(\{\begin{array}{l}y=\frac{2}{3}x-2 \\ y=-\frac{1}{3}x-5\end{array}\)

Try it.

\(\{\begin{array}{l}x+y=-4 \\ -x+2y=-2\end{array}\)

Solution

\((-2,-2)\)

Try it.

\(\{\begin{array}{l}-x+3y=3 \\ x+3y=3\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+3y=3 \\ x+3y=12\end{array}\)

Solution

\((3,3)\)

Try it.

\(\{\begin{array}{l}2x-y=4 \\ 2x+3y=12\end{array}\)

Try it.

\(\{\begin{array}{l}x+3y=-6 \\ y=-\frac{4}{3}x+4\end{array}\)

Solution

\((6,-4)\)

Try it.

\(\{\begin{array}{l}-x+2y=-6 \\ y=-\frac{1}{2}x-1\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+4y=4 \\ y=\frac{1}{2}x\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}3x+5y=10 \\ y=-\frac{3}{5}x+1\end{array}\)

Try it.

\(\{\begin{array}{l}\ 4x-3y=8 \\ 8x-6y=14\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}x+3y=4 \\ -2x-6y=3\end{array}\)

Try it.

\(\{\begin{array}{l}x=-3y+4 \\ 2x+6y=8\end{array}\)

Solution

infinite solutions

Try it.

\(\{\begin{array}{l}4x=3y+7 \\ 8x-6y=14\end{array}\)

Try it.

\(\{\begin{array}{l}2x+\ y=6 \\ -8x-4y=-24\end{array}\)

Solution

infinite solutions

Try it.

\(\{\begin{array}{l}5x+\ 2y=7 \\ -10x-4y=-14\end{array}\)

Without graphing, determine the number of solutions and then classify the system of equations.

Try it.

\(\{\begin{array}{l}y=\frac{2}{3}x+1 \\ -2x+3y=5\end{array}\)

Solution

No solution, Inconsistent, Independent

Try it.

\(\{\begin{array}{l}y=\frac{3}{2}x+1 \\ 2x-3y=7\end{array}\)

Try it.

\(\{\begin{array}{l}5x+3y=4 \\ 2x-3y=5\end{array}\)

Solution

1 point, consistent and independent

Try it.

\(\{\begin{array}{l}y=-\frac{1}{2}x+5 \\ x+2y=10\end{array}\)

Try it.

\(\{\begin{array}{l}5x-2y=10 \\ y=\frac{5}{2}x-5\end{array}\)

Solution

infinite solutions, consistent, dependent

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

Try it.

\(\{\begin{array}{l}2x+y=-4 \\ 3x-2y=-6\end{array}\)

Try it.

\(\{\begin{array}{l}2x+y=-2 \\ 3x-y=7\end{array}\)

Solution

\((1,-4)\)

Try it.

\(\{\begin{array}{l}x-2y=-5 \\ 2x-3y=-4\end{array}\)

Try it.

\(\{\begin{array}{l}x-3y=-9 \\ 2x+5y=4\end{array}\)

Solution

\((-3,2)\)

Try it.

\(\{\begin{array}{l}5x-2y=-6 \\ y=3x+3\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+2y=6 \\ y=-3x+1\end{array}\)

Solution

\((-\frac{1}{2},\frac{5}{2})\)

Try it.

\(\{\begin{array}{l}2x+5y=1 \\ y=\frac{1}{3}x-2\end{array}\)

Try it.

\(\{\begin{array}{l}3x+4y=1 \\ y=-\frac{2}{5}x+2\end{array}\)

Solution

\((-5,4)\)

Try it.

\(\{\begin{array}{l}2x+y=5 \\ x-2y=-15\end{array}\)

Try it.

\(\{\begin{array}{l}4x+y=10 \\ x-2y=-20\end{array}\)

Solution

\((0,10)\)

Try it.

\(\{\begin{array}{l}y=-2x-1 \\ y=-\frac{1}{3}x+4\end{array}\)

Try it.

\(\{\begin{array}{l}y=x-6 \\ y=-\frac{3}{2}x+4\end{array}\)

Solution

\((4,-2)\)

Try it.

\(\{\begin{array}{l}\ \ x=2y \\ 4x-8y=0\end{array}\)

Try it.

\(\{\begin{array}{l}\ 2x-16y=8 \\ -x-8y=-4\end{array}\)

Solution

\((4,0)\)

Try it.

\(\{\begin{array}{l}y=\frac{7}{8}x+4 \\ -7x+8y=6\end{array}\)

Try it.

\(\{\begin{array}{l}y=-\frac{2}{3}x+5 \\ 2x+3y=11\end{array}\)

Solution

no solution

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

Try it.

\(\{\begin{array}{l}5x+2y=2 \\ -3x-y=0\end{array}\)

Try it.

\(\{\begin{array}{l}6x-5y=-1 \\ 2x+y=13\end{array}\)

Solution

\((4,5)\)

Try it.

\(\{\begin{array}{l}2x-5y=7 \\ 3x-y=17\end{array}\)

Try it.

\(\{\begin{array}{l}5x-3y=-1 \\ 2x-y=2\end{array}\)

Solution

\((7,12)\)

Try it.

\(\{\begin{array}{l}3x-5y=-9 \\ 5x+2y=16\end{array}\)

Try it.

\(\{\begin{array}{l}4x-3y=3 \\ 2x+5y=-31\end{array}\)

Solution

\((-3,-5)\)

Try it.

\(\{\begin{array}{l}3x+8y=-3 \\ 2x+5y=-3\end{array}\)

Try it.

\(\{\begin{array}{l}11x+9y=-5 \\ 7x+5y=-1\end{array}\)

Solution

\((2,-3)\)

Try it.

\(\{\begin{array}{l}3x+8y=67 \\ 5x+3y=60\end{array}\)

Try it.

\(\{\begin{array}{l}2x+9y=-4 \\ 3x+13y=-7\end{array}\)

Solution

\((-11,2)\)

Try it.

\(\{\begin{array}{l}\frac{1}{3}x-y=-3 \\ x+\frac{5}{2}y=2\end{array}\)

Try it.

\(\{\begin{array}{l}x+\frac{1}{2}y=\frac{3}{2} \\ \frac{1}{5}x-\frac{1}{5}y=3\end{array}\)

Solution

\((6,-9)\)

Try it.

\(\{\begin{array}{l}x+\frac{1}{3}y=-1 \\ \frac{1}{3}x+\frac{1}{2}y=1\end{array}\)

Try it.

\(\{\begin{array}{l}\frac{1}{3}x-y=-3 \\ \frac{2}{3}x+\frac{5}{2}y=3\end{array}\)

Solution

\((-3,2)\)

Try it.

\(\{\begin{array}{l}2x+y=3 \\ 6x+3y=9\end{array}\)

Try it.

\(\{\begin{array}{l}x-4y=-1 \\ -3x+12y=3\end{array}\)

Solution

infinitely many

Try it.

\(\{\begin{array}{l}-3x-y=8 \\ 6x+2y=-16\end{array}\)

Try it.

\(\{\begin{array}{l}4x+3y=2 \\ 20x+15y=10\end{array}\)

Solution

infinitely many

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

Try it.

ⓐ \(\{\begin{array}{l}8x-15y=-32 \\ 6x+3y=-5\end{array}\)
ⓑ \(\{\begin{array}{l}x=4y-3 \\ 4x-2y=-6\end{array}\)

Try it.

ⓐ \(\{\begin{array}{l}y=7x-5 \\ 3x-2y=16\end{array}\)
ⓑ \(\{\begin{array}{l}12x-5y=-42 \\ 3x+7y=-15\end{array}\)

Solution

ⓐ substitution ⓑ elimination

Try it.

ⓐ \(\{\begin{array}{l}y=4x+9 \\ 5x-2y=-21\end{array}\)
ⓑ \(\{\begin{array}{l}9x-4y=24 \\ 3x+5y=-14\end{array}\)

Try it.

ⓐ \(\{\begin{array}{l}14x-15y=-30 \\ 7x+2y=10\end{array}\)
ⓑ \(\{\begin{array}{l}x=9y-11 \\ 2x-7y=-27\end{array}\)

Solution

ⓐ elimination ⓑ substituion

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For the equation \(y=\frac{2}{3}x-4,\)
ⓐ Is \((6,0)\) a solution? ⓑ Is \((-3,-2)\) a solution?
If you missed this problem, review .

జవాబు వెల్లడి చేయండి

ⓐ yes; ⓑ no
Find the slope and y-intercept of the line \(3x-y=12.\)
If you missed this problem, review .

జవాబు వెల్లడి చేయండి

\(m=3;\ b=-12\)
Find the x- and y-intercepts of the line \(2x-3y=12.\)
If you missed this problem, review .

జవాబు వెల్లడి చేయండి

\(\left(6,\ 0\right),\ \left(0,\ -4\right)\)
Determine whether the ordered pair is a solution to the system \(\{\begin{array}{l}x-y=-1 \\ 2x-y=-5\end{array}.\)
ⓐ \((-2,-1)\) ⓑ \((-4,-3)\)

జవాబు వెల్లడి చేయండి

ⓐ
ⓑ
Determine whether the ordered pair is a solution to the system \(\begin{array}{l}\{\begin{array}{l}3x+y=0 \\ x+2y=-5\end{array}. \\ \end{array}\)
ⓐ \((1,-3)\) ⓑ \((0,0)\)

జవాబు వెల్లడి చేయండి

ⓐ yes ⓑ no
Determine whether the ordered pair is a solution to the system \(\begin{array}{l}\{\begin{array}{l}x-3y=-8 \\ -3x-y=4\end{array}. \\ \end{array}\)
ⓐ \((2,-2)\) ⓑ \((-2,2)\)

జవాబు వెల్లడి చేయండి

ⓐ no ⓑ yes
Solve the system by graphing \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)
Solve the system by graphing: \(\{\begin{array}{l}x-3y=-3 \\ x+y=5\end{array}.\)

జవాబు వెల్లడి చేయండి

\((3,2)\)
Solve the system by graphing: \(\{\begin{array}{l}-x+y=1 \\ 3x+2y=12\end{array}.\)

జవాబు వెల్లడి చేయండి

\((2,3)\)

Solve the system by graphing: \(\{\begin{array}{l}3x+y=-1 \\ 2x+y=0\end{array}.\)

జవాబు వెల్లడి చేయండి

We’ll solve both of these equations for \(y\) so that we can easily graph them using their slopes and y-intercepts.


Solve the first equation for y.
Find the slope and y-intercept.
Solve the second equation for y.
Find the slope and y-intercept.
Graph the lines.
Determine the point of intersection.	The lines intersect at \((-1,2).\)
Check the solution in both equations.
	The solution is \((-1,2).\)

Solve the system by graphing: \(\{\begin{array}{l}-x+y=1 \\ 2x+y=10\end{array}.\)

జవాబు వెల్లడి చేయండి

\((3,4)\)
Solve the system by graphing: \(\{\begin{array}{l}2x+y=6 \\ x+y=1\end{array}.\)

జవాబు వెల్లడి చేయండి

\((5,-4)\)

Solve the system by graphing: \(\{\begin{array}{l}y=\frac{1}{2}x-3 \\ x-2y=4\end{array}.\)

జవాబు వెల్లడి చేయండి


To graph the first equation, we will use its slope and y-intercept.
To graph the second equation, we will use the intercepts.

Graph the lines.
Determine the points of intersection.	The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.

Solve the system by graphing: \(\{\begin{array}{l}y=-\frac{1}{4}x+2 \\ x+4y=-8\end{array}.\)

జవాబు వెల్లడి చేయండి

no solution
Solve the system by graphing: \(\{\begin{array}{l}y=3x-1 \\ 6x-2y=6\end{array}.\)

జవాబు వెల్లడి చేయండి

no solution

Solve the system by graphing: \(\{\begin{array}{l}y=2x-3 \\ -6x+3y=-9\end{array}.\)

జవాబు వెల్లడి చేయండి


Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.

Graph the lines.
	The lines are the same! Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to this system.

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y-intercept.

Solve the system by graphing: \(\{\begin{array}{l}y=-3x-6 \\ 6x+2y=-12\end{array}.\)

జవాబు వెల్లడి చేయండి

infinitely many solutions
Solve the system by graphing: \(\{\begin{array}{l}y=\frac{1}{2}x-4 \\ 2x-4y=16\end{array}.\)

జవాబు వెల్లడి చేయండి

infinitely many solutions

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ \(\{\begin{array}{l}y=3x-1 \\ 6x-2y=12\end{array}\) ⓑ \(\{\begin{array}{l}2x+y=-3 \\ x-5y=5\end{array}\)

జవాబు వెల్లడి చేయండి

ⓐ We will compare the slopes and intercepts of the two lines.

The first equation is already in slope-intercept form.	\(\ \begin{array}{lllll}\{\begin{array}{lll}y & = & 3x-1 \\ 6x-2y & = & 12\end{array} \\ y\ =\ 3x-1\end{array}\)
Write the second equation in slope-intercept form.	\(\ \begin{array}{llllllll} & & & & & 6x-2y & = & 12 \\ & & & & & -2y & = & -6x+12 \\ & & & & & \frac{-2y}{-2} & = & \frac{-6x+12}{-2} \\ & & & & & y & = & 3x-6\end{array}\)
Find the slope and intercept of each line.	\(\begin{array}{llllll}y & = & 3x-1 & \ y & = & 3x-6 \\ m & = & 3 & \ m & = & 3 \\ b & = & -1 & \ b & = & -6\end{array}\)
	Since the slopes are the same and y-intercepts are different, the lines are parallel.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

	\(\begin{array}{lllll}\{\begin{array}{lll}2x+y & = & -3 \\ x-5y & = & 5\end{array}\end{array}\)
Write both equations in slope–intercept form.	\(\ \begin{array}{llllllll}2x+y & = & -3 & & & x-5y & = & 5 \\ y & = & -2x-3 & & & -5y & = & \text{-}x+5 \\ & & & & & \frac{-5y}{-5} & = & \frac{-x+5}{-5} \\ & & & & & y & = & \frac{1}{5}x-1\end{array}\)
Find the slope and intercept of each line.	\(\ \begin{array}{llllll}y & = & -2x-3 & \ y & = & \frac{1}{5}x-1 \\ m & = & -2 & \ m & = & \frac{1}{5} \\ b & = & -3 & \ b & = & -1\end{array}\)
	Since the slopes are different, the lines intersect.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ \(\{\begin{array}{l}y=-2x-4 \\ 4x+2y=9\end{array}\) ⓑ \(\{\begin{array}{l}3x+2y=2 \\ 2x+y=1\end{array}\)

జవాబు వెల్లడి చేయండి

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ \(\{\begin{array}{l}y=\frac{1}{3}x-5 \\ x-3y=6\end{array}\) ⓑ \(\{\begin{array}{l}x+4y=12 \\ -x+y=3\end{array}\)

జవాబు వెల్లడి చేయండి

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Solve the system by substitution: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)
Solve the system by substitution: \(\{\begin{array}{l}-2x+y=-11 \\ x+3y=9\end{array}.\)

జవాబు వెల్లడి చేయండి

\((6,1)\)
Solve the system by substitution: \(\{\begin{array}{l}2x+y=-1 \\ 4x+3y=3\end{array}.\)

జవాబు వెల్లడి చేయండి

\((-3,5)\)

Solve the system by substitution: \(\{\begin{array}{l}4x+2y=4 \\ 6x-y=8\end{array}.\)

జవాబు వెల్లడి చేయండి

We need to solve one equation for one variable. We will solve the first equation for y.


Solve the first equation for y. Substitute \(-2x+2\) for y in the second equation.
Replace the y with \(-2x+2.\)
Solve the equation for x.
Substitute \(x=\frac{5}{4}\) into \(4x+2y=4\) to find y.
	The ordered pair is \((\frac{5}{4},-\frac{1}{2}).\)
Check the ordered pair in both equations.
	The solution is \((\frac{5}{4},-\frac{1}{2}).\)

Solve the system by substitution: \(\{\begin{array}{l}x-4y=-4 \\ -3x+4y=0\end{array}.\)

జవాబు వెల్లడి చేయండి

\((2,\frac{3}{2})\)
Solve the system by substitution: \(\{\begin{array}{l}4x-y=0 \\ 2x-3y=5\end{array}.\)

జవాబు వెల్లడి చేయండి

\((-\frac{1}{2},-2)\)
Solve the system by elimination: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)
Solve the system by elimination: \(\{\begin{array}{l}3x+y=5 \\ 2x-3y=7\end{array}.\)

జవాబు వెల్లడి చేయండి

\((2,-1)\)
Solve the system by elimination: \(\{\begin{array}{l}4x+y=-5 \\ -2x-2y=-2\end{array}.\)

జవాబు వెల్లడి చేయండి

\((-2,3)\)

Solve the system by elimination: \(\{\begin{array}{l}4x-3y=9 \\ 7x+2y=-6\end{array}.\)

జవాబు వెల్లడి చేయండి

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.


Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x.
Substitute \(x=0\) into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is \((0,-3).\)
Check that the ordered pair is a solution to both original equations.
	The solution is \((0,-3).\)

Solve the system by elimination: \(\{\begin{array}{l}3x-4y=-9 \\ 5x+3y=14\end{array}.\)

జవాబు వెల్లడి చేయండి

\((1,3)\)
Solve each system by elimination: \(\{\begin{array}{l}7x+8y=4 \\ 3x-5y=27\end{array}.\)

జవాబు వెల్లడి చేయండి

\((4,-3)\)

Solve the system by elimination: \(\{\begin{array}{l}x+\frac{1}{2}y=6 \\ \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\end{array}.\)

జవాబు వెల్లడి చేయండి

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.


To clear the fractions, multiply each equation by its LCD.
Simplify.
Now we are ready to eliminate one of the variables. Notice that both equations are in standard form.
We can eliminate \(y\) by multiplying the top equation by \(-4.\)
Simplify and add. Substitute \(x=3\) into one of the original equations.
Solve for \(y\).
Write the solution as an ordered pair.	The ordered pair is \((3,6).\)
Check that the ordered pair is a solution to both original equations.
	The solution is \((3,6).\)

Solve each system by elimination: \(\{\begin{array}{l}\frac{1}{3}x-\frac{1}{2}y=1 \\ \frac{3}{4}x-y=\frac{5}{2}\end{array}.\)

జవాబు వెల్లడి చేయండి

\((6,2)\)
Solve each system by elimination: \(\{\begin{array}{l}x+\frac{3}{5}y=-\frac{1}{5} \\ -\frac{1}{2}x-\frac{2}{3}y=\frac{5}{6}\end{array}.\)

జవాబు వెల్లడి చేయండి

\((1,-2)\)

Solve the system by elimination: \(\{\begin{array}{l}3x+4y=12 \\ y=3-\frac{3}{4}x\end{array}.\)

జవాబు వెల్లడి చేయండి

	\(\{\begin{array}{l}3x+4y=12 \\ y=3-\frac{3}{4}x\end{array}\)
Write the second equation in standard form.	\(\{\begin{array}{l}3x+4y=12 \\ \frac{3}{4}x+y=3\end{array}\)
Clear the fractions by multiplying the second equation by 4.	\(\{\begin{array}{l}3x+4y=12 \\ 4(\frac{3}{4}x+y)=4(3)\end{array}\)
Simplify.	\(\{\begin{array}{l}3x+4y=12 \\ 3x+4y=12\end{array}\)
To eliminate a variable, we multiply the second equation by −1. Simplify and add.	\(\begin{array}{l}\underset{______________}{\{\begin{array}{l}\ 3x+4y=12 \\ -3x-4y=-12\end{array}} \\ \\ 0=0\end{array}\)

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Solve the system by elimination: \(\{\begin{array}{l}5x-3y=15 \\ y=-5+\frac{5}{3}x\end{array}.\)

జవాబు వెల్లడి చేయండి

infinitely many solutions
Solve the system by elimination: \(\{\begin{array}{l}x+2y=6 \\ y=-\frac{1}{2}x+3\end{array}.\)

జవాబు వెల్లడి చేయండి

infinitely many solutions
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ \(\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\) ⓑ \(\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\)

జవాబు వెల్లడి చేయండి

ⓐ
\[\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\]
Since both equations are in standard form, using elimination will be most convenient.
ⓑ
\[\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\]
Since one equation is already solved for y, using substitution will be most convenient.

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Linear Equations with Two Variables

Determine whether an ordered pair is a solution of a system of equations
Solve a system of linear equations by graphing
Solve a system of equations by substitution
Solve a system of equations by elimination
Choose the most convenient method to solve a system of linear equations
Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

మీ సొంత ప్రయత్నించండి

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ఇంకా Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value