Solutions to Problem Set 7

Math 101-06

9-18-2017

[Lines]

1. Determine whether the following lines are parallel, perpendicular, or neither.

$$3 x - 4 y + 7 = 0 \quad\hbox{and}\quad 8 x + 6 y = 5.$$

$$\eqalign{ 3 x - 4 y + 7 & = 0 \cr 3 x + 7 & = 4 y \cr \noalign{\vskip2pt} \dfrac{3}{4} x + \dfrac{7}{4} & = y \cr}$$

The slope of the line is $\dfrac{3}{4}$ .

$$\eqalign{ 8 x + 6 y & = 5 \cr 6 y & = -8 x + 5 \cr \noalign{\vskip2pt} y & = -\dfrac{4}{3} + \dfrac{5}{6} \cr}$$

The slope of the line is $-\dfrac{4}{3}$ .

Since $\dfrac{3}{4}$ and $-\dfrac{4}{3}$ are negative reciprocals, the lines are perpendicular.


2. Determine whether the following lines are parallel, perpendicular, or neither.

$$y = -5 x + 7 \quad\hbox{and}\quad 30 x + 6 y = 3.$$

The slope of $y = -5 x + 7$ is -5.

$$\eqalign{ 30 x + 6 y & = 3 \cr 6 y & = -30 x + 3 \cr \noalign{\vskip2pt} y & = -5 x + \dfrac{1}{2} \cr}$$

The slope is -5.

Since both lines have slope -5, the lines are parallel.


3. Find the equation of the line which passes through the point $(4, -8)$ and is parallel to the line $7 x + y = 3$ .

$$\eqalign{ 7 x + y & = 3 \cr y & = -7 x + 3 \cr \noalign{\vskip2pt} y & = -7 x + 3 \cr}$$

The slope is -7.

A parallel line must also have slope -7. Using this slope and the point $(4, -8)$ , the equation is

$$y + 8 = -7(x - 4), \quad\hbox{or}\quad y = -7 x + 20.\quad\halmos$$


4. Find the equation of the line which passes through the point $(2, 1)$ and is perpendicular to the line $-3 x + 5 y = 8$ .

$$\eqalign{ -3 x + 5 y & = 8 \cr 5 y & = 3 x + 8 \cr \noalign{\vskip2pt} y & = \dfrac{3}{5} x + \dfrac{8}{5} \cr}$$

The slope is $\dfrac{3}{5}$ .

A perpendicular line must have slope $-\dfrac{5}{3}$ . Using this slope and the point $(2, 1)$ , the equation is

$$y - 1 = -\dfrac{5}{3} (x - 2), \quad\hbox{or}\quad y = -\dfrac{5}{3} x + \dfrac{13}{3}.\quad\halmos$$


[Systems]

5. Solve the following system of equations for x and y.

$$\eqalign{ 3 x + 5 y & = 11 \cr 3 x - 5 y & = 1 \cr}$$ The coefficients of x are the same, so subtract the equations to eliminate x:

$$\eqalign{ 3 x + 5 y & = 11 \cr 3 x - 5 y & = 1 \cr \noalign{\vskip2pt \hrule \vskip2pt} 10 y & = 10 \cr y & = 1 \cr}$$

Returning to the original equations, the coefficients of y are the same, but opposite in sign, so add the equations to eliminate y:

$$\eqalign{ 3 x + 5 y & = 11 \cr 3 x - 5 y & = 1 \cr \noalign{\vskip2pt \hrule \vskip2pt} 6 x & = 12 \cr x & = 2 \cr}$$

The solution is $x = 2$ and $y
   = 1$ .


6. Solve the following system of equations for x and y.

$$\eqalign{ 2 x + 3 y & = -4 \cr 5 x + 8 y & = 7 \cr}$$ Multiply the first equation by 5, multiply the second equation by 2, then subtract the resulting equations:

$$\eqalign{ 10 x + 15 y & = -20 \cr 10 x + 16 y & = 14 \cr \noalign{\vskip2pt \hrule \vskip2pt} -y & = -34 \cr y & = 34 \cr}$$

Returning to the original equations, multiply the first equation by 8, multiply the second equation by 3, then subtract the resulting equations:

$$\eqalign{ 16 x + 24 y & = -32 \cr 15 x + 24 y & = 21 \cr \noalign{\vskip2pt \hrule \vskip2pt} x & = -53 \cr}$$

The solution is $x = -53$ and $y = 34$ .


7. Solve the following system of equations for x and y.

$$\eqalign{ 7 x + 2 y & = 1 \cr -x + 3 y & = 5 \cr}$$ Multiply the first equation by 3, multiply the second equation by 2, then subtract the second equation from the first equation:

$$\eqalign{ 21 x + 6 y & = 3 \cr -2 x + 6 y & = 10 \cr \noalign{\vskip2pt \hrule \vskip2pt} 23 x & = -7 \cr \noalign{\vskip2pt} x & = -\dfrac{7}{23} \cr}$$

Multiply the second equation by 7 and add the first equation:

$$\eqalign{ 7 x + 2 y & = 1 \cr -7 x + 21 y & = 35 \cr \noalign{\vskip2pt \hrule \vskip2pt} 23 y & = 36 \cr \noalign{\vskip2pt} y & = \dfrac{36}{23} \cr}$$

The solution is $x =
   -\dfrac{7}{23}$ and $y = \dfrac{36}{23}$ .


8. Solve the following system of equations for x and y.

$$\eqalign{ x - 2 y & = 3 \cr -3 x + 6 y & = 1 \cr}$$ Multiply the first equation by 3 and add it to the second equation:

$$\eqalign{ 3 x - 6 y & = 9 \cr -3 x + 6 y & = 1 \cr \noalign{\vskip2pt \hrule \vskip2pt} 0 & = 10 \cr}$$

The last equation is false. Therefore, the system has no solutions.


To think is not enough; you must think of something. - Jules Renard


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