Problems with Systems of Equations

The exercises with systems of equations can be solved using various methods. However, the most commonly used ones are the substitution method and the elimination method. In this article, we will solve various systems of two equations using these two methods.

We will start with a quick summary of these methods and then look at several worked-out examples.

ALGEBRA
linear systems of equations

Relevant for

Learning how to solve exercises of systems of equations.

See examples

ALGEBRA
linear systems of equations

Relevant for

Learning how to solve exercises of systems of equations.

See examples

Systems of equations and solutions

Systems of equations are sets of two or more equations, which have to be solved simultaneously. In this article, we will focus entirely on systems of two equations with two unknowns and we will solve exercises only with these systems.

Solution to a system

For an ordered pair to be a solution to a system of equations, it has to satisfy both equations at the same time. If an ordered pair is a solution to one equation but not the other, then it is not a solution to the system of equations.

A system that has at least one solution is called a consistent system.

A system that has no solutions is called an inconsistent system.

There are three possible results that we can find in the systems of equations:

1. One solution: this happens when the lines that represent the equations intersect at a single point.

2. No solution: this happens when the lines that represent the equations are parallel.

3. Infinite solutions: this happens when the line of one equation overlaps the line of the other equation.


Solve systems with the substitution method

Follow these steps to solve the systems of equations problems with the substitution method:

Step 1: Simplify if possible.

Remove parentheses, combine like terms, and eliminate fractions.

Step 2: Solve an equation for one variable.

It doesn’t matter which equation we choose or for which variable we solve.

Step 3: Substitute the expression from step 2 into the other equation.

By doing this, we will obtain an equation with only one unknown.

Step 4: Solve for the variable we have remaining.

If you need help with this, you can look at our guide on how to solve equations with one unknown.

Step 5: Solve for the other variable.

We substitute the value obtained in step 4 into any equation and solve.

Step 6: Check the solution in both equations.

We plug the values into the equations and check the result.

PROBLEM 1

Solve with the substitution method: \left\{ {\begin{array}{*{20}{c}} {5x-y=2} \\ {y=4x-1} \end{array}} \right..

Step 1: Simplify if possible.

Both equations are already simplified.

Step 2: Solve an equation for one variable.

The second equation is already solved for the variable y:

y=4x-1

Step 3: Substitute the expression from step 2 into the other equation.

We substitute the expression y=4x-1 in the first equation: 

5x-y=2

5x-(4x-1)=2

5x-4x+1=2

Step 4: Solve for the remaining variable:

5x-4x+1=2

x=1

Step 5: Solve for the second variable.

We substitute x=1 in the second equation: 

y=4x-1

y=4(1)-1

y=3

Step 6: Check the solution in both equations.

PROBLEM 2

Solve with the substitution method: \left\{ {\begin{array}{*{20}{c}} {3x+y=11} \\ {2x-y=-1} \end{array}} \right.

Step 1: Simplify if possible.

Both equations are already simplified.

Step 2: Solve an equation for one variable.

We solve the second equation for y:

2x-y=-1

y=2x+1

Step 3: Substitute the expression from step 2 into the other equation.

We substitute the expression y=2x+1 n the first equation: 

3x+y=11

3x+2x+1=11

Step 4: Solve for the remaining variable:

3x+2x+1=11

5x=10

x=2

Step 5: Solve for the second variable.

We substitutex=2 in the second equation: 

2x-y=-1

2(2)-y=-1

4-y=-1

y=5

Step 6: Check the solution in both equations.

PROBLEM 3

Solve with the substitution method: \left\{ {\begin{array}{*{20}{c}} {4x-2y=15} \\ {3x+2y=25} \end{array}} \right.

Step 1: Simplify if possible.

Both equations are already simplified.

Step 2: Solve an equation for one variable.

We solve the second equation for y:

3x+2y=25

2y=25-3x

y=\frac{{25}}{2}-\frac{3}{2}x

Step 3: Substitute the expression from step 2 into the other equation.

We substitute the expression y=\frac{{25}}{2}-\frac{3}{2}x in the first equation: 

4x-2y=15

4x-2(\frac{{25}}{2}-\frac{3}{2}x)=15

4x-25+3x=15

Step 4: Solve for the remaining variable:

4x-25+3x=15

7x=35

x=5

Step 5: Solve for the second variable.

We substitute x=5 in the second equation: 

3x+2y=25

3(5)+2y=25

2y=25-15

y=5

Step 6: Check the solution in both equations.

PROBLEM 4

Solve with the substitution method: \left\{ {\begin{array}{*{20}{c}} {6x+3y=-9} \\ {2x-2y=-12} \end{array}} \right.

Step 1: Simplify if possible.

We can simplify the first equation by dividing it by 3: 2x+y=-3.

We can simplify the second equation by dividing it by 2: x-y=-6.

Step 2: Solve an equation for one variable.

We solve the first equation for y:

2x+y=-3

y=-3-2x

Step 3: Substitute the expression from step 2 into the other equation.

We substitute the expression y=-3-2x in the second equation: 

x-y=-6

x-(-3-2x)=-6

x+3+2x=-6

Step 4: Solve for the remaining variable:

x+3+2x=-6

3x=-9

x=-3

Step 5: Solve for the second variable.

We substitute x=-3 in the second equation: 

x-y=-6

-3-y=-6

-y=-3

y=3

Step 6: Check the solution in both equations.

Try solving the following practice problems

Solve the system 4x-y=4, y=x+2.

Choose an answer






Solve the system -3x+y=6, 2x-y=-3.

Choose an answer







Solve systems with the elimination method

Follow these steps to solve the systems of equations problems with the substitution method:

Step 1: Simplify if possible and put the equations in the form Ax + By = C.

Step 2: Multiply one or both equations by a number to create opposite coefficients for either x or y.

We are going to add the equations and we have to make one of the variables disappear. We can achieve this with opposite coefficients since their sum is 0.

For example, if we have 3x in the first equation and 2x in the second equation, we can multiply the first by -2 and the second by 3, thus we get -6 in the first and 6 in the second.

Step 3: Add the equations.

One of the variables will be eliminated and we will obtain an equation with only one unknown.

Step 4: Solve for the remaining variable.

If you need help with this, you can look at our guide on how to solve equations with one unknown.

Step 5: Solve for the second variable.

Substitute the value you found in step 4 into either of the two equations and solve for the other variable.

Step 6: Check the solution in both equations.

PROBLEM 1

Solve with elimination method: \left\{ {\begin{array}{*{20}{c}} {3x+3y=15} \\ {-3x+y=-3} \end{array}} \right.

Step 1: Simplify if possible and write in the form Ax+By=C.

Both equations are already simplified and in the form Ax+By=C.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We already have opposite coefficients in the variable x.

Step 3: Add the equations. 

3x+3y=15

+  \hspace{1cm}    -3x+y=-3               

___________________

4y=12

Step 4: Solve for the remaining variable:

4y=12

y=3

Step 5: Solve for the second variable.

We substitute y=3 in the first equation: 

3x+3y=15

3x+3(3)=15

3x+9=15

3x=6

x=2

Step 6: Check the solution in both equations.

PROBLEM 2

Solve with elimination method: \left\{ {\begin{array}{*{20}{c}} {-2x+y=9} \\ {-3x-2y=3} \end{array}} \right.

Step 1: Simplify if possible and write in the form Ax+By=C.

Both equations are already simplified and in the form Ax+By=C.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We multiply the first equation by 2 to eliminate the y:

\left\{ {\begin{array}{*{20}{c}} {-4x+2y=18} \\ {-3x-2y=3} \end{array}} \right.

Step 3: Add the equations. 

-4x+2y=18

+  \hspace{1cm}    -3x-2y=3               

___________________

-7x=21

Step 4: Solve for the remaining variable:

-7x=21

x=-3

Step 5: Solve for the second variable.

We substitute x=-3 in the first equation: 

-2x+y=9

-2(-3)+y=9

6+y=9

y=3

Step 6: Check the solution in both equations.

PROBLEM 3

Solve with the elimination method: \left\{ {\begin{array}{*{20}{c}} {x=4y+5} \\ {2(x+2)+3y=3} \end{array}} \right.

Step 1: Simplify if possible and place in the form Ax+By=C:

\left\{ {\begin{array}{*{20}{c}} {x-4y=5} \\ {2x+3y=1} \end{array}} \right.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We multiply the first equation by -2 to eliminate the x:

\left\{ {\begin{array}{*{20}{c}} {-2x+8y=-10} \\ {2x+3y=1} \end{array}} \right.

Step 3: Add the equations. 

-2x+8y=10

+  \hspace{1cm}    2x+3y=1               

___________________

11x=11

Step 4: Solve for the remaining variable:

11x=11

x=1

Step 5: Solve for the second variable.

We substitutex=1 in the first equation: 

x=4y+5

1=4y+5

-4y=4

y=-1

Step 6: Check the solution in both equations.

PROBLEM 4

Solve with elimination method: \left\{ {\begin{array}{*{20}{c}} {2x+2y=20} \\ {-3x+3y=0} \end{array}} \right.

Step 1: Simplify if possible and write in the form Ax+By=C.

We can simplify both equations: we divide the first by 2 and the second by 3:

\left\{ {\begin{array}{*{20}{c}} {x+y=10} \\ {-x+y=0} \end{array}} \right.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We already have opposite coefficients in the x:

Step 3: Add the equations. 

x+y=10

+  \hspace{1cm}    -x+y=0               

___________________

2y=10

Step 4: Solve for the remaining variable:

2y=10

y=5

Step 5: Solve for the second variable.

We substitutey=5 in the second equation: 

-x+y=0

-x+5=0

x=5

Step 6: Check the solution in both equations.

Try solving the following practice problems

Solve the system -2x-2y=2, x+2y=1.

Choose an answer






Solve the system 3x-2y=7, 2x+y=14.

Choose an answer







See also

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