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 $\begin{array}{l} {\begin{cases} 2 x - y = 1 \\ 4 x - 2 y = 4 \end{cases} & \begin{array}{l} (1) \\ (2) \end{array} \end{array}$

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

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

Step 3: Solve the equation obtained in step 2.
$\begin{array}{l} 4 x - 2 (2 x - 1) & = & 4 \\ 4 x - 4 x + 2 & = & 4 \\ 2 & \neq & 4 \end{array}$

Computations have eliminated all the variables and produce a contradiction. These lines are parallel.
A graph of two parallel lines. One line is labeled with the equation two x minus y is equal to one and passes through the points one, one, and zero, negative one. A second line is labeled with the equation four x minus two y is equal to four and passes through the points one, zero, and zero, negative two.
This system is inconsistent.

Practice set b

Slove the system ${\begin{cases} 7 x - 3 y = 2 \\ 14 x - 6 y = 1 \end{cases}$

Substitution produces $4 \neq 1,$ or $\frac{1}{2} \neq 2$ , 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 $\begin{array}{l} {\begin{cases} 4 x + 8 y = 8 \\ 3 x + 6 y = 6 \end{cases} & \begin{array}{l} (1) \\ (2) \end{array} \end{array}$

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

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

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

Computations have eliminated all the variables and produced an identity. These lines are coincident.
A graph of two coincident lines. The line is labeled with the equation x plus two y is equal to two and a second label with the equation three x plus six y is equal to six. The lines pass through the points zero, one and two, zero. Since the lines are coincident, they have the same graph.
This system is dependent.

Practice set c

Solve the system ${\begin{cases} 4 x + 3 y = 1 \\ - 8 x - 6 y = - 2 \end{cases}$

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

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

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

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

Step 4: Substitute $x = \frac{9}{17}$ into the equation obtained in step $1, y = \frac{- 3}{2} x + \frac{1}{2} .$
$y = \frac{- 3}{2} (\frac{9}{17}) + \frac{1}{2}$
$y = \frac{- 27}{34} + \frac{17}{34} = \frac{- 10}{34} = \frac{- 5}{17}$
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 $(\frac{9}{17}, \frac{- 5}{17}) .$

Practice set d

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

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

Exercises

For the following problems, solve the systems by substitution.

${\begin{array}{l} 3 x + 2 y & = & 9 \\ y & = & - 3 x + 6 \end{array}$

$(1, 3)$

${\begin{array}{l} 5 x - 3 y & = & - 6 \\ y & = & - 4 x + 12 \end{array}$

${\begin{array}{l} 2 x + 2 y & = & 0 \\ x & = & 3 y - 4 \end{array}$

$(- 1, 1)$

${\begin{array}{l} 3 x + 5 y & = & 9 \\ x & = & 4 y - 14 \end{array}$

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

$(2, 2)$

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

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

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

${\begin{array}{l} - x - y & = & 5 \\ 2 x + y & = & 5 \end{array}$

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

Dependent (same line)

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

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

$(1, - 2)$

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

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

inconsistent (parallel lines)

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

${\begin{array}{l} x + y & = & - 6 \\ x - y & = & 4 \end{array}$

$(- 1, - 5)$

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

${\begin{array}{l} 4 x - 2 y & = & 7 \\ y & = & 4 \end{array}$

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

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

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

$(- 1, - 3)$

${\begin{array}{l} y + 7 x + 4 & = & 0 \\ x = - 7 y & + & 28 \end{array}$

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

$(4, - 1)$

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

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

inconsistent (parallel lines)

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

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

$(1, - 1)$

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

Exercises for review

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

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

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

( [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 $2 x^{2} + 2 x - 3 = 0.$

( [link] ) Solve by graphing ${\begin{cases} x - y = 1 \\ 2 x + y = 5 \end{cases}$
An xy coordinate plane with gridlines labeled negative five and five with increments of one unit for both axes.

$(2, 1)$
A graph of two lines intersecting at a point with coordinates negative two, one. One of the lines is passing through a point with coordinates zero, five and the other line is passing through a point with coordinates zero, negative one.

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Source: OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3

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