4.1 Solving systems of linear equations by substitution  (Page 2/2)

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Substitution and parallel lines

If computations eliminate all the variables and produce a contradiction, the two lines of a system are parallel, and the system is called inconsistent.

Sample set b

Solve the system $\begin{array}{lll}\left\{\begin{array}{l}2x-y=1\\ 4x-2y=4\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$

Step 1:  Solve equation 1 for $y.$
$\begin{array}{lll}\hfill 2x-y& =\hfill & 1\hfill \\ \hfill -y& =\hfill & -2x+1\hfill \\ \hfill y& =\hfill & 2x-1\hfill \end{array}$

Step 2:  Substitute the expression $2x-1$ for $y$ into equation 2.
$4x-2\left(2x-1\right)=4$

Step 3:  Solve the equation obtained in step 2.
$\begin{array}{lll}4x-2\left(2x-1\right)\hfill & =\hfill & 4\hfill \\ \hfill 4x-4x+2& =\hfill & 4\hfill \\ \hfill 2& \ne \hfill & 4\hfill \end{array}$

Computations have eliminated all the variables and produce a contradiction. These lines are parallel.

This system is inconsistent.

Practice set b

Slove the system $\left\{\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7x-3y=2\\ 14x-6y=1\end{array}$

Substitution produces $4\ne 1,$ or $\frac{1}{2}\ne 2$ , a contradiction. These lines are parallel and the system is inconsistent.

Substitution and coincident lines

The following rule alerts us to the fact that the two lines of a system are coincident.

Substitution and coincident lines

If computations eliminate all the variables and produce an identity, the two lines of a system are coincident and the system is called dependent.

Sample set c

Solve the system $\begin{array}{lll}\left\{\begin{array}{l}4x+8y=8\\ 3x+6y=6\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$

Step 1:  Divide equation 1 by 4 and solve for $x.$
$\begin{array}{lll}\hfill 4x+8y& =\hfill & 8\hfill \\ \hfill x+2y& =\hfill & 2\hfill \\ \hfill x& =\hfill & -2y+2\hfill \end{array}$

Step 2:  Substitute the expression $-2y+2$ for $x$ in equation 2.
$3\left(-2y+2\right)+6y=6$

Step 3:  Solve the equation obtained in step 2.
$\begin{array}{lll}\hfill 3\left(-2y+2\right)+6y& =\hfill & 6\hfill \\ \hfill -6y+6+6y& =\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\hfill \end{array}$

Computations have eliminated all the variables and produced an identity. These lines are coincident.

This system is dependent.

Practice set c

Solve the system $\left\{\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x+3y=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ -8x-6y=-2\end{array}$

Computations produce $-2=-2,$ an identity. These lines are coincident and the system is dependent.

Systems in which a coefficient of one of the variables is not 1 or cannot be made to be 1 without introducing fractions are not well suited for the substitution method. The problem in Sample Set D illustrates this “messy” situation.

Sample set d

Solve the system $\begin{array}{lll}\left\{\begin{array}{l}3x+2y=1\\ 4x-3y=3\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$

Step 1:  We will solve equation $\left(1\right)$ for $y.$
$\begin{array}{lll}\hfill 3x+2y& =\hfill & 1\hfill \\ \hfill 2y& =\hfill & -3x+1\hfill \\ \hfill y& =\hfill & \frac{-3}{2}x+\frac{1}{2}\hfill \end{array}$

Step 2:  Substitute the expression $\frac{-3}{2}x+\frac{1}{2}$ for $y$ in equation $\left(2\right).$
$4x-3\left(\frac{-3}{2}x+\frac{1}{2}\right)=3$

Step 3:  Solve the equation obtained in step 2.
$\begin{array}{lllll}\hfill 4x-3\left(\frac{-3}{2}x+\frac{1}{2}\right)& =\hfill & 3\hfill & \hfill & \text{Multiply\hspace{0.17em}both\hspace{0.17em}sides\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}LCD},\text{\hspace{0.17em}2}.\hfill \\ \hfill 4x+\frac{9}{2}x-\frac{3}{2}& =\hfill & 3\hfill & \hfill & \hfill \\ \hfill 8x+9x-3& =\hfill & 6\hfill & \hfill & \hfill \\ \hfill 17x-3& =\hfill & 6\hfill & \hfill & \hfill \\ \hfill 17x& =\hfill & 9\hfill & \hfill & \hfill \\ \hfill x& =\hfill & \frac{9}{17}\hfill & \hfill & \hfill \end{array}$

Step 4:  Substitute $x=\frac{9}{17}$ into the equation obtained in step $1,\text{\hspace{0.17em}}y=\frac{-3}{2}x+\frac{1}{2}.$
$y=\frac{-3}{2}\left(\frac{9}{17}\right)+\frac{1}{2}$
$y=\frac{-27}{34}+\frac{17}{34}=\frac{-10}{34}=\frac{-5}{17}$
We now have $x=\frac{9}{17}$ and $y=\frac{-5}{17}.$

Step 5:  Substitution will show that these values of $x$ and $y$ check.

Step 6:  The solution is $\left(\frac{9}{17},\frac{-5}{17}\right).$

Practice set d

Solve the system $\left\{\begin{array}{l}9x-5y=-4\\ 2x+7y=-9\end{array}$

These lines intersect at the point $\left(-1,-1\right).$

Exercises

For the following problems, solve the systems by substitution.

$\left\{\begin{array}{lll}3x+2y\hfill & =\hfill & 9\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\hfill & =\hfill & -3x+6\hfill \end{array}$

$\left(1,3\right)$

$\left\{\begin{array}{lll}5x-3y\hfill & =\hfill & -6\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\hfill & =\hfill & -4x+12\hfill \end{array}$

$\left\{\begin{array}{lll}2x+2y\hfill & =\hfill & 0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}x\hfill & =\hfill & 3y-4\hfill \end{array}$

$\left(-1,1\right)$

$\left\{\begin{array}{lll}3x+5y\hfill & =\hfill & 9\hfill \\ \text{\hspace{0.17em}}x\hfill & =\hfill & 4y-14\hfill \end{array}$

$\left\{\begin{array}{lll}-3x+y\hfill & =\hfill & -4\hfill \\ 2x+3y\hfill & =\hfill & 10\hfill \end{array}$

$\left(2,2\right)$

$\left\{\begin{array}{lll}-4x+y\hfill & =\hfill & -7\hfill \\ 2x+5y\hfill & =\hfill & 9\hfill \end{array}$

$\left\{\begin{array}{lll}6x-6\hfill & =\hfill & 18\hfill \\ x+3y\hfill & =\hfill & 3\hfill \end{array}$

$\left(4,-\frac{1}{3}\right)$

$\left\{\begin{array}{lll}-x-y\hfill & =\hfill & 5\hfill \\ 2x+y\hfill & =\hfill & 5\hfill \end{array}$

$\left\{\begin{array}{lll}-5x+y\hfill & =\hfill & 4\hfill \\ 10x-2y\hfill & =\hfill & -8\hfill \end{array}$

Dependent (same line)

$\left\{\begin{array}{lll}x+4y\hfill & =\hfill & 1\hfill \\ -3x-12y\hfill & =\hfill & -1\hfill \end{array}$

$\left\{\begin{array}{lll}4x-2y\hfill & =\hfill & 8\hfill \\ 6x+3y\hfill & =\hfill & 0\hfill \end{array}$

$\left(1,-2\right)$

$\left\{\begin{array}{lll}2x+3y\hfill & =\hfill & 12\hfill \\ 2x+4y\hfill & =\hfill & 18\hfill \end{array}$

$\left\{\begin{array}{lll}3x-9y\hfill & =\hfill & 6\hfill \\ 6x-18y\hfill & =\hfill & 5\hfill \end{array}$

inconsistent (parallel lines)

$\left\{\begin{array}{lll}-x+4y\hfill & =\hfill & 8\hfill \\ 3x-12y\hfill & =\hfill & 10\hfill \end{array}$

$\left\{\begin{array}{lll}x+y\hfill & =\hfill & -6\hfill \\ x-y\hfill & =\hfill & 4\hfill \end{array}$

$\left(-1,-5\right)$

$\left\{\begin{array}{lll}2x+y\hfill & =\hfill & 0\hfill \\ x-3y\hfill & =\hfill & 0\hfill \end{array}$

$\left\{\begin{array}{lll}4x-2y\hfill & =\hfill & 7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\hfill & =\hfill & 4\hfill \end{array}$

$\left(\frac{15}{4},4\right)$

$\left\{\begin{array}{lll}x+6y\hfill & =\hfill & 11\hfill \\ x\hfill & =\hfill & -1\hfill \end{array}$

$\left\{\begin{array}{lll}2x-4y\hfill & =\hfill & 10\hfill \\ 3x=5y\hfill & +\hfill & 12\hfill \end{array}$

$\left(-1,-3\right)$

$\left\{\begin{array}{lll}y+7x+4\hfill & =\hfill & 0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}x=-7y\hfill & +\hfill & 28\hfill \end{array}$

$\left\{\begin{array}{l}x+4y=0\\ x+\frac{2}{3}y=\frac{10}{3}\end{array}$

$\left(4,-1\right)$

$\left\{\begin{array}{l}x=24-5y\\ x-\frac{5}{4}y=\frac{3}{2}\end{array}$

$\left\{\begin{array}{l}x=11-6y\\ 3x+18y=-33\end{array}$

inconsistent (parallel lines)

$\left\{\begin{array}{l}2x+\frac{1}{3}y=4\\ 3x+6y=39\end{array}$

$\left\{\begin{array}{l}\frac{4}{5}x+\frac{1}{2}y=\frac{3}{10}\\ \frac{1}{3}x+\frac{1}{2}y=\frac{-1}{6}\end{array}$

$\left(1,-1\right)$

$\left\{\begin{array}{lll}x-\frac{1}{3}y\hfill & =\hfill & \frac{-8}{3}\hfill \\ -3x+y\hfill & =\hfill & 1\hfill \end{array}$

Exercises for review

( [link] ) Find the quotient: $\begin{array}{l}\frac{{x}^{2}-x-12}{{x}^{2}-2x-15}\hfill \end{array}÷\frac{{x}^{2}-3x-10}{{x}^{2}-2x-8}.$

$\frac{{\left(x-4\right)}^{2}}{{\left(x-5\right)}^{2}}$

( [link] ) Find the difference: $\begin{array}{l}\frac{x+2}{{x}^{2}+5x+6}\hfill \end{array}-\frac{x+1}{{x}^{2}+4x+3}.$

( [link] ) Simplify $-\sqrt{81{x}^{8}{y}^{5}{z}^{4}.}$

$-9{x}^{4}{y}^{2}{z}^{2}\sqrt{y}$

( [link] ) Use the quadratic formula to solve $2{x}^{2}+2x-3=0.$

( [link] ) Solve by graphing $\left\{\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-y=1\\ 2x+y=5\end{array}$

$\left(2,1\right)$

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