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

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<H2>Solution of a linear system of two equations in two unknowns by the Substitution method</H2>

This lesson describes the &nbsp;<B><U>Substitution method</U></B>&nbsp; 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. &nbsp;In this way you reduce the original 
system of two equations with two unknowns to the single equation with one unknown. &nbsp;When this reduction is done, &nbsp;simplify the obtained equation and solve it.

After completing this, &nbsp;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.

<H3>Example 1</H3>

Solve a system of equations

{{{system (2x +  y =  5,
          -4x + 6y = -2
)}}}

Express &nbsp;{{{y}}}&nbsp; from the first equation:

{{{y = 5 - 2x}}}.

Substitute this to the second equation:

{{{-4x +6(5-2x) = -2}}}.

Simplify the last equation by collecting common terms and solve it for &nbsp;{{{x}}}:

{{{-4x + 30 - 12x = -2}}},
{{{-4x - 12x = -2 - 30}}},
{{{-16x = -32}}},
{{{16x = 32}}},
{{{x = 2}}}.

Back-substitute this value of &nbsp;{{{x}}}&nbsp; to the first original equation:

{{{2*2 + y = 5}}}.

Simplify and solve this equation for &nbsp;{{{y}}}:

{{{y = 5-4 = 1}}}.

Thus, &nbsp;the solution of the original system of equations is &nbsp;{{{x = 2}}}, {{{y=1}}}.

Now, &nbsp;check the calculated solution. &nbsp;Simply substitute the found values of &nbsp;{{{x=2}}}&nbsp; and &nbsp;{{{y=1}}}&nbsp; into the original equations. &nbsp;You will get

{{{2*2+1 = 5}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;for the left side of the first equation, and this is identical to its right side;
{{{-4*2+6*1 = -8+6 = -2}}} &nbsp;&nbsp;for the left side of the second equation, &nbsp;and this is identical to its right side.

The check shows that the solution is correct.

<H3>Example 2</H3>

Solve a system of equations

{{{system (3x - 2y =  5,
           5x -  y =  6
)}}}

Express &nbsp;{{{y}}}&nbsp; from the second equation:

{{{y = -6 + 5x}}}.

Substitute this to the first equation:

{{{3x - 2(-6 + 5x) = 5}}}.

Simplify the last equation by collecting common terms and solve it for &nbsp;{{{x}}}:

{{{3x - 10x) = 5 - 12}}}.
{{{-7x) = -7}}}.
{{{x = 1}}}.

The solution of the last equation is &nbsp;{{{x = 1}}}.

Back-substitute this value of &nbsp;{{{x}}}&nbsp; to the expression &nbsp;{{{y = -6 + 5x}}}:

{{{y = -6 + 5*1 = -6 + 5 = -1}}}.

The solution for &nbsp;{{{y}}}&nbsp; is &nbsp;{{{y = -1}}}.

Now, &nbsp;check the calculated solution. Simply substitute the found values of &nbsp;{{{x=1}}}&nbsp; and &nbsp;{{{y=-1}}}&nbsp; to the original equations. &nbsp;You will get

{{{3*1-2*(-1) = 5}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;for the left side of the first equation, and this is identical to its right side;
{{{5*1-(-1) =5+1 = 6}}} &nbsp;&nbsp;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 &nbsp;<B>Examples 1</B>&nbsp; and &nbsp;<B>2</B>&nbsp; are the cases, &nbsp;when the linear equation system has the unique solution.
These are examples of the &nbsp;<B>consistent</B>&nbsp; equation systems with &nbsp;<B>independent</B>&nbsp; equations.

<H3>Example 3</H3>

Solve a system of equations

{{{system (3x - 2y =  5,
          12x - 8y = 20
)}}}

Express &nbsp;{{{y}}}&nbsp; from the first equation:

{{{y = (-5 + 3x)/2}}}.

Substitute this expression for &nbsp;{{{y}}}&nbsp; to the second equation:

{{{12x - 8*(-5+3x)/2 = 20}}}, &nbsp;or, &nbsp;step by step simplifying,
{{{12x -12x + 20 = 20}}}.
{{{0x = 0}}}.

The last equation has infinitely many solutions: &nbsp;each and any value of &nbsp;{{{x}}}&nbsp; satisfies to this equation.

Since the last equation does not produce any restriction to &nbsp;{{{x}}}, &nbsp;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 = (-5+2y)/3}}}

is the solution. &nbsp;For example, &nbsp;the following pairs of variables values are solutions:

{{{x=-5/3}}}, {{{y=0}}},
{{{x=-1}}}, {{{y=1}}},
{{{x=-1/3}}}, {{{y=2}}},
{{{x=1/3}}}, {{{y=3}}},
{{{x=1}}}, {{{y=4}}}, ... .

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

<H3>Example 4</H3>

Solve a system of equations

{{{system (3x - 2y =  5,
          12x - 8y = 15
)}}}

Express &nbsp;{{{y}}}&nbsp; from the first equation. &nbsp;You will get:

{{{y = (-5 + 3x)/2}}}.

Substitute this expression for &nbsp;{{{y}}}&nbsp; to the second equation. &nbsp;You will get:

{{{12x - 8*(-5+3x)/2 = 15}}}, &nbsp;or, &nbsp;step by step simplifying,
{{{12x -12x + 20 = 15}}}.
{{{0x = 5}}}.

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

So, &nbsp;the original system of equations has no solutions.

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

<H3><U>Summary</U></H3>
The &nbsp;<B>Substitution method</B>&nbsp; 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. &nbsp;In this way you reduce the original system of two equations with two unknowns to the single equation with 
one unknown. &nbsp;When this reduction is done, &nbsp;simplify the obtained equation and solve it.

After completing this, &nbsp;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 &nbsp;<B>Substitution method</B>, &nbsp;you may have one of the three following typical cases:

Case 1. &nbsp;The system of equations is &nbsp;<B>consistent</B>&nbsp; and the equations are &nbsp;<B>independent</B>.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The system of equations has the solution in this case and it is unique.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The substitution procedure works smoothly in this case both at the direct-substitution and back-substitution steps.

Case 2. &nbsp;The system of equations is &nbsp;<B>consistent</B>, &nbsp;but the equations are &nbsp;<B>dependent</B>.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The system of equations has infinitely many solutions in this case.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Typically the substitution procedure leads to the equation &nbsp;(identity)&nbsp; of the form &nbsp;{{{0*x = 0}}}&nbsp; or &nbsp;{{{0*y = 0}}}&nbsp; that have infinitely many solutions.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This is the case when the coefficients and the right side of one equation are proportional to that of the second equation.

Case 3. &nbsp;The system of equations is &nbsp;<B>inconsistent</B>, &nbsp;the equations are &nbsp;<B>dependent</B>.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The system of equations has no solutions in this case.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Typically the substitution procedure leads to the unsolvable equation of the form &nbsp;{{{0*x = 1}}}&nbsp; or &nbsp;{{{0*y = 1}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This is the case when the coefficients of one equation are proportional to that of the second equation, &nbsp;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 &nbsp;(Algebra-I curriculum)&nbsp; are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF = http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-syst-of-two-eqns-with-two-unknowns-Elimination-method.lesson>Solution of a linear system of two equations in two unknowns by the Elimination method</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/coordinate/Solution-of-the-lin-syst-of-two-eqns-with-two-unknowns-using-det.lesson>Solution of a linear system of two equations in two unknowns using determinant</A>

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