Lesson Solution of a linear system of two equations in two unknowns by the Substitution method

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Solution of a linear system of two equations in two unknowns by the Substitution method


This lesson describes the  Substitution method  for solving a linear system of two equations in two unknowns.
The method is to express one variable via another using one equation and then to substitute this expression into another equation.  In this way you reduce the original
system of two equations with two unknowns to the single equation with one unknown.  When this reduction is done,  simplify the obtained equation and solve it.
After completing this,  back-substitute the found value to either of the original equations and find the value of the remaining variable.

Examples below show how this method works.

Example 1


Solve a system of equations

system+%282x+%2B++y+=++5%2C%0D%0A++++++++++-4x+%2B+6y+=+-2%0D%0A%29

Express  y  from the first equation:
y+=+5+-+2x.

Substitute this to the second equation:
-4x+%2B6%285-2x%29+=+-2.

Simplify the last equation by collecting common terms and solve it for  x:
-4x+%2B+30+-+12x+=+-2,
-4x+-+12x+=+-2+-+30,
-16x+=+-32,
16x+=+32,
x+=+2.

Back-substitute this value of  x  to the first original equation:
2%2A2+%2B+y+=+5.

Simplify and solve this equation for  y:
y+=+5-4+=+1.

Thus,  the solution of the original system of equations is  x+=+2, y=1.

Now,  check the calculated solution.  Simply substitute the found values of  x=2  and  y=1  into the original equations.  You will get
2%2A2%2B1+=+5                      for the left side of the first equation, and this is identical to its right side;
-4%2A2%2B6%2A1+=+-8%2B6+=+-2   for the left side of the second equation,  and this is identical to its right side.

The check shows that the solution is correct.

Example 2


Solve a system of equations

system+%283x+-+2y+=++5%2C%0D%0A+++++++++++5x+-++y+=++6%0D%0A%29

Express  y  from the second equation:
y+=+-6+%2B+5x.

Substitute this to the first equation:
3x+-+2%28-6+%2B+5x%29+=+5.

Simplify the last equation by collecting common terms and solve it for  x:
3x+-+10x%29+=+5+-+12.
-7x%29+=+-7.
x+=+1.

The solution of the last equation is  x+=+1.

Back-substitute this value of  x  to the expression  y+=+-6+%2B+5x:
y+=+-6+%2B+5%2A1+=+-6+%2B+5+=+-1.

The solution for  y  is  y+=+-1.

Now,  check the calculated solution. Simply substitute the found values of  x=1  and  y=-1  to the original equations.  You will get
3%2A1-2%2A%28-1%29+=+5         for the left side of the first equation, and this is identical to its right side;
5%2A1-%28-1%29+=5%2B1+=+6   for the left side of the second equation, and this is identical to its right side.

The check shows that the solution is correct.

Note that  Examples 1  and  2  are the cases,  when the linear equation system has the unique solution.
These are examples of the  consistent  equation systems with  independent  equations.

Example 3


Solve a system of equations

system+%283x+-+2y+=++5%2C%0D%0A++++++++++12x+-+8y+=+20%0D%0A%29

Express  y  from the first equation:

y+=+%28-5+%2B+3x%29%2F2.

Substitute this expression for  y  to the second equation:
12x+-+8%2A%28-5%2B3x%29%2F2+=+20,  or,  step by step simplifying,
12x+-12x+%2B+20+=+20.
0x+=+0.

The last equation has infinitely many solutions:  each and any value of  x  satisfies to this equation.

Since the last equation does not produce any restriction to  x,  the original system of equations is equivalent to the single equation
3x+-+2y+=++5,
which has infinitely many solutions: for every value of y the value of
x+=+%28-5%2B2y%29%2F3
is the solution.  For example,  the following pairs of variables values are solutions:
x=-5%2F3, y=0,
x=-1, y=1,
x=-1%2F3, y=2,
x=1%2F3, y=3,
x=1, y=4, ... .

In this example,  the linear equation system has infinitely many solutions,  because the coefficients and the right side of one equation are proportional
to that of the second equation. This is the case of the  consistent  equation system with  dependent  equations.

Example 4


Solve a system of equations

system+%283x+-+2y+=++5%2C%0D%0A++++++++++12x+-+8y+=+15%0D%0A%29

Express  y  from the first equation.  You will get:

y+=+%28-5+%2B+3x%29%2F2.

Substitute this expression for  y  to the second equation.  You will get:
12x+-+8%2A%28-5%2B3x%29%2F2+=+15,  or,  step by step simplifying,
12x+-12x+%2B+20+=+15.
0x+=+5.

The last equation has no solutions because its left side is always equal to zero,  while the right side is not equal to zero.

So,  the original system of equations has no solutions.

In this example,  the linear equation system has no solutions, because the coefficients of one equation are proportional to that of the second equation,  while
the right sides are out of this proportionality.  This is the case of the  inconsistent  equation system with  dependent  equations.

Summary


The  Substitution method  for solving a system of two linear equations with two unknowns is to express one variable via another using one equation and then
to substitute this expression into another equation.  In this way you reduce the original system of two equations with two unknowns to the single equation with
one unknown.  When this reduction is done,  simplify the obtained equation and solve it.
After completing this,  back-substitute the found value to either one of the original equations and find the value of the remaining variable.

When solving a linear system equations by the  Substitution method,  you may have one of the three following typical cases:

Case 1.  The system of equations is  consistent  and the equations are  independent.
            The system of equations has the solution in this case and it is unique.
            The substitution procedure works smoothly in this case both at the direct-substitution and back-substitution steps.

Case 2.  The system of equations is  consistent,  but the equations are  dependent.
            The system of equations has infinitely many solutions in this case.
            Typically the substitution procedure leads to the equation  (identity)  of the form  0%2Ax+=+0  or  0%2Ay+=+0  that have infinitely many solutions.
            This is the case when the coefficients and the right side of one equation are proportional to that of the second equation.

Case 3.  The system of equations is  inconsistent,  the equations are  dependent.
            The system of equations has no solutions in this case.
            Typically the substitution procedure leads to the unsolvable equation of the form  0%2Ax+=+1  or  0%2Ay+=+1.
            This is the case when the coefficients of one equation are proportional to that of the second equation,  but the right sides are out of this proportionality.


My other lessons on solving linear systems of two equations in two unknowns in this site  (Algebra-I curriculum)  are
    - Solution of a linear system of two equations in two unknowns by the Elimination method
    - Solution of a linear system of two equations in two unknowns using determinant
    - Geometry interpretation of a linear system of two equations in two unknowns
    - Useful tricks when solving systems of 2 equations in 2 unknowns by the Substitution method
    - Solving word problems using linear systems of two equations in two unknowns

    - Word problems that lead to a simple system of two equations in two unknowns
    - Oranges and grapefruits
    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
    - Using systems of equations to solve problems on shares
    - Using systems of equations to solve problems on investment
    - Two mechanics work on a car
    - The Robinson family and the Sanders family each used their sprinklers last summer
    - Roses and violets
    - Counting calories and grams of fat in combined food
    - A theater group made appearances in two cities
    - Exchange problems solved using systems of linear equations
    - Typical word problems on systems of 2 equations in 2 unknowns
    - HOW TO algebraize and solve these problems on 2 equations in 2 unknowns
    - One unusual problem to solve using system of two equations
    - Non-standard problems with a tricky setup
    - Sometimes one equation is enough to find two unknowns in a unique way
    - Solving mentally word problems on two equations in two unknowns
    - Word problem to solve combined system of linear equations and a price equation
    - Solving systems of non-linear equations by reducing to linear ones
    - Solving word problems for 3 unknowns by reducing to equations in 2 unknowns
    - System of equations helps to solve a problem for the Thanksgiving day

    - OVERVIEW of lessons on solving systems of two linear equations in two unknowns

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