Exercises: systems with two variables

Determine the solution sets of the following systems of nonlinear equations.

$\begin{array}{rcll} (I) & \frac{4}{x} + \frac{8}{y} & = & \frac{5}{3} \\ (II) & \frac{2}{x} - \frac{4}{y} & = & - \frac{1}{6} \end{array}$
where $x, y \neq 0$

For this task you need the following basic knowledge: System of linear equations
You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.
Here the addition method is suitable. If you transform the two equations so that the terms with $x$ or $y$ are identical, you can subtract the two transformed equations from each other and only have to solve one equation with one variable.
Solution by addition method
Using the addition method you get to the following solution:
Given is: $\begin{array}{rcll} (I) & \frac{4}{x} + \frac{8}{y} & = & \frac{5}{3} \\ (II) & \frac{2}{x} - \frac{4}{y} & = & - \frac{1}{6} \end{array}$
The equation $(II)$ contains a term $- \frac{4}{y}$ and equation $(I)$ is a multiple of $- \frac{4}{y}$ , namely $\frac{8}{y}$ . So you can multiply $(II)$ by $2$ and add both equations to eliminate $\frac{8}{y}$ .
$(II) \frac{2}{x} - \frac{4}{y}$ $=$ $- \frac{1}{6}$ $\cdot 2$
$({II}^{'}) \frac{4}{x} - \frac{8}{y}$ $=$ $- \frac{2}{6}$
So the new system of equations is:
$\begin{array}{rcll} (I) & \frac{4}{x} + \frac{8}{y} & = & \frac{5}{3} \\ ({II}^{'}) & \frac{4}{x} - \frac{8}{y} & = & - \frac{2}{6} \end{array}$
$(I) + ({II}^{'})$ :
$\begin{array}{rcll} (I) & \frac{4}{x} + \frac{8}{y} & = & \frac{5}{3} \\ + ({II}^{'}) & \frac{4}{x} - \frac{8}{y} & = & - \frac{2}{6} \end{array}$
$\frac{8}{x} = \frac{5}{3} - \frac{2}{6}$
Now solve the new equation for $x$ .
$\frac{8}{x}$ $=$ $\frac{5}{3} - \frac{2}{6}$
$\frac{8}{x}$ $=$ $\frac{5}{3} - \frac{1}{3}$
$\frac{8}{x}$ $=$ $\frac{4}{3}$
↓
invert the fractions
$\frac{x}{8}$ $=$ $\frac{3}{4}$ $\cdot 8$
$x$ $=$ $\frac{3 \cdot 8}{4}$
$x$ $=$ $6$
Now, plug $x$ into one of the above equations, e.g. $(I)$ , and solve for $y$ .
$(I) \frac{4}{x} + \frac{8}{y}$ $=$ $\frac{5}{3}$
↓
plug in $x = 6$
$\frac{4}{6} + \frac{8}{y}$ $=$ $\frac{5}{3}$ $- \frac{4}{6}$
$\frac{8}{y}$ $=$ $\frac{5}{3} - \frac{4}{6}$
$\frac{8}{y}$ $=$ $\frac{5}{3} - \frac{2}{3}$
$\frac{8}{y}$ $=$ $\frac{3}{3}$
$\frac{8}{y}$ $=$ $1$ $\cdot y$
$y$ $=$ $8$
Write down the solution set. $L = {(x | y)} = {(6 | 8)}$ .
Try to solve these non-linear equations using the methods you already know for linear systems of equations.
$\begin{array}{rrcl} (I) & \frac{7}{x} - \frac{12}{y} & = & \frac{5}{6} \\ (II) & \frac{4}{y} + \frac{5}{2} & = & \frac{9}{x} \end{array}$
where $x, y \neq 0$

For this task you need the following basic knowledge: System of linear equations
You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.
Here the addition method is suitable. If you transform the two equations so that the terms with $x$ or $y$ are identical, you can subtract the two transformed equations from each other and only have to solve one equation with one variable.
Solution by the addition method
Using the addition method you get to the following solution:
Given is: $\begin{array}{rrcl} (I) & \frac{7}{x} - \frac{12}{y} & = & \frac{5}{6} \\ (II) & \frac{4}{y} + \frac{5}{2} & = & \frac{9}{x} \end{array}$
Transform $(II)$ such that on one side you have only variables and on the other side only numbers.
$(II) \frac{4}{y} + \frac{5}{2}$ $=$ $\frac{9}{x}$ $- \frac{9}{x} - \frac{5}{2}$
$({II}^{'}) - \frac{9}{x} + \frac{4}{y}$ $=$ $- \frac{5}{2}$
The new system of equations is now:
$\begin{array}{rrcl} (I) & \frac{7}{x} - \frac{12}{y} & = & \frac{5}{6} \\ ({II}^{'}) & - \frac{9}{x} + \frac{4}{y} & = & - \frac{5}{2} \end{array}$
Now, equation $({II}^{'})$ contains the term $\frac{4}{y}$ and equation $(I)$ is a multiple of $\frac{4}{y}$ , namely $- \frac{12}{y}$ . So you multiply $({II}^{'})$ by $3$ and add both equations in order to eliminate $\frac{12}{y}$ .
$({II}^{'}) - \frac{9}{x} + \frac{4}{y}$ $=$ $- \frac{5}{2}$ $\cdot 3$
$({II}^{''}) - \frac{27}{x} + \frac{12}{y}$ $=$ $- \frac{15}{2}$
$(I) + ({II}^{''})$ :
$(I) \frac{7}{x} - \frac{12}{y} = \frac{5}{6}$
$(I I) - \frac{27}{x} + \frac{12}{y} = - \frac{15}{2}$
$\begin{array}{rrcr} (I) & \frac{7}{x} - \frac{12}{y} & = & \frac{5}{6} \\ + ({II}^{''}) & - \frac{27}{x} + \frac{12}{y} & = & - \frac{15}{2} \end{array}$
$- \frac{20}{x} = - \frac{40}{6}$
Now, you solve $- \frac{20}{x} = - \frac{40}{6}$ , by inverting the fraction.
$- \frac{20}{x}$ $=$ $- \frac{40}{6}$
$- \frac{x}{20}$ $=$ $- \frac{6}{40}$ $\cdot (- 20)$
$x$ $=$ $\frac{6 \cdot 20}{2 \cdot 20}$
↓
shorten
$x$ $=$ $\frac{6}{2}$
$x$ $=$ $3$
Now, plug $x$ into one of the above equations, for instance $(II)$ , and solve for $y$ .
$(II) \frac{4}{y} + \frac{5}{2}$ $=$ $\frac{9}{x}$
↓
plug in $x = 3$
$\frac{4}{y} + \frac{5}{2}$ $=$ $\frac{9}{3}$ $- \frac{5}{2}$
$\frac{4}{y}$ $=$ $\frac{9}{3} - \frac{5}{2}$
$\frac{4}{y}$ $=$ $\frac{18}{6} - \frac{15}{6}$
$\frac{4}{y}$ $=$ $\frac{3}{6}$
$\frac{4}{y}$ $=$ $\frac{1}{2}$
↓
invert the fraction
$\frac{y}{4}$ $=$ $\frac{2}{1}$ $\cdot 4$
$y$ $=$ $8$
Write down the solution set. $L = {(x | y)} = {(3 | 8)}$
Try to solve these non-linear equations using the methods you already know for linear systems of equations.
$\begin{array}{l} (I) \frac{4}{3 x + 1} = \frac{2}{3 y - 13} \\ (II) \frac{2}{5 x - 10} = \frac{4}{7 y - 6} \end{array}$
where $x \notin {- \frac{1}{3}; 2}$ and $y \notin {\frac{13}{3}; \frac{6}{7}}$

For this task you need the following basic knowledge: System of linear equations
This non-linear system of equations can be transformed into a linear system of equations. And you may solve linear systems using the substitution method, the addition method or by equating coefficients.
Given is: $\begin{array}{l} (I) \frac{4}{3 x + 1} = \frac{2}{3 y - 13} \\ (II) \frac{2}{5 x - 10} = \frac{4}{7 y - 6} \end{array}$
Transform $(I)$ and $(II)$ , such that both equations no longer contain a fraction.
$(I) \frac{4}{3 x + 1}$ $=$ $\frac{2}{3 y - 13}$ $\cdot (3 x + 1) \cdot (3 y - 13)$
↓
multiply by the denominators
$4 \cdot (3 y - 13)$ $=$ $2 \cdot (3 x + 1)$
$(I^{'}) 12 y - 52$ $=$ $6 x + 2$
$(II) \frac{2}{5 x - 10}$ $=$ $\frac{4}{7 y - 6}$ $\cdot (5 x - 10) \cdot (7 y - 6)$
↓
multiply by the denominators
$2 \cdot (7 y - 6)$ $=$ $4 \cdot (5 x - 10)$
$({II}^{'}) 14 y - 12$ $=$ $20 x - 40$
Thus, the linear system of equations is:
$\begin{array}{l} (I^{'}) 12 y - 52 = 6 x + 2 \\ ({II}^{'}) 14 y - 12 = 20 x - 40 \end{array}$
Now you can use the known procedures for solving the system of equations.
Solution by the substitution method
With the substitution method you get this solution:
Given is: $\begin{array}{l} (I^{'}) 12 y - 52 = 6 x + 2 \\ ({II}^{'}) 14 y - 12 = 20 x - 40 \end{array}$
Transform $(I^{'})$ , such that $x$ appears only one one side.
$(I^{'}) 12 y - 52$ $=$ $6 x + 2$ $- 2$
$12 y - 54$ $=$ $6 x$ $: 6$
$(I^{''}) 2 y - 9$ $=$ $x$
Plug $x = 2 y - 9$ into $({II}^{'})$ .
$({II}^{'}) 14 y - 12$ $=$ $20 x - 40$
↓
plug in $x = 2 y - 9$
$14 y - 12$ $=$ $20 \cdot (2 y - 9) - 40$
↓
Solve for $y$
$14 y - 12$ $=$ $40 y - 180 - 40$
$14 y - 12$ $=$ $40 y - 220$ $- 14 y + 220$
$208$ $=$ $26 y$ $: 26$
$y$ $=$ $8$
Plug $y = 8$ into $(I^{'})$ , in order to find $x$ .
$(I^{'}) x$ $=$ $2 y - 9$
↓
plug in $y = 8$
$=$ $2 \cdot 8 - 9$
$=$ $16 - 9$
$=$ $7$
Write down the solution set. $L = {(x | y)} = {(7 | 8)}$
Additionally: Check your solution
Plug $x = 7$ and $y = 8$ into equations $(I)$ and $(II)$ and check if the equation is satisfied.
$(I) \frac{4}{3 x + 1}$ $=$ $\frac{2}{3 y - 13}$
↓
plug in $x = 7$ and $y = 8$
$\frac{4}{3 \cdot 7 + 1}$ $=$ $\frac{2}{3 \cdot 8 - 13}$
$\frac{4}{21 + 1}$ $=$ $\frac{2}{24 - 13}$
$\frac{4}{22}$ $=$ $\frac{2}{11}$
$\frac{2}{11}$ $=$ $\frac{2}{11} ✓$
$(II) \frac{2}{5 x - 10}$ $=$ $\frac{4}{7 y - 6}$
↓
plug in $x = 7$ and $y = 8$
$\frac{2}{5 \cdot 7 - 10}$ $=$ $\frac{4}{7 \cdot 8 - 6}$
$\frac{2}{35 - 10}$ $=$ $\frac{4}{56 - 6}$
$\frac{2}{25}$ $=$ $\frac{4}{50}$
$\frac{2}{25}$ $=$ $\frac{2}{25} ✓$
Both equations are satisfied, so our solution is correct.
Try to solve these non-linear equations using the methods you already know for linear systems of equations.
$\begin{array}{rrcl} (I) & \frac{3}{2 x - 1} - \frac{8}{3 y + 2} & = & - \frac{1}{5} \\ (II) & \frac{5}{2 x - 1} + \frac{4}{3 y + 2} & = & \frac{8}{15} \end{array}$
where $x \neq \frac{1}{2}$ and $y \neq - \frac{2}{3}$

For this task you need the following basic knowledge: System of linear equations
You may solve linear systems of equations using the substitution method, the addition method or by equating coefficients. Such methods can also be effective for some non-linear system of equations.
Here the addition method is suitable. If you transform the two equations so that the terms with $x$ or $y$ are identical, you can subtract the two transformed equations from each other and only have to solve one equation with one variable.
Solution using the addition method
Given is: $\begin{array}{rrcl} (I) & \frac{3}{2 x - 1} - \frac{8}{3 y + 2} & = & - \frac{1}{5} \\ (II) & \frac{5}{2 x - 1} + \frac{4}{3 y + 2} & = & \frac{8}{15} \end{array}$
The equation $(II)$ contains the term $\frac{4}{3 y + 2}$ and $(I)$ contains a multiple of $\frac{4}{3 y + 2}$ , namely $- \frac{8}{3 y + 2}$ . So you multiply $(II)$ by $2$ and add both equations in order to eliminate $\frac{8}{3 y + 2}$ .
$(II) \frac{5}{2 x - 1} + \frac{4}{3 y + 2}$ $=$ $\frac{8}{15}$ $\cdot 2$
$({II}^{'}) \frac{10}{2 x - 1} + \frac{8}{3 y + 2}$ $=$ $\frac{16}{15}$
So the new system of equations is:
$\begin{array}{rrcl} (I) & \frac{3}{2 x - 1} - \frac{8}{3 y + 2} & = & - \frac{1}{5} \\ ({II}^{'}) & \frac{10}{2 x - 1} + \frac{8}{3 y + 2} & = & \frac{16}{15} \end{array}$
$(I) + ({II}^{'})$ :
$\begin{array}{rrcl} (I) & \frac{3}{2 x - 1} - \frac{8}{3 y + 2} & = & - \frac{1}{5} \\ + ({II}^{'}) & \frac{10}{2 x - 1} + \frac{8}{3 y + 2} & = & \frac{16}{15} \end{array}$
$\frac{13}{2 x - 1} = - \frac{1}{5} + \frac{16}{15}$
Now solve the equation resulting from the addition for $x$ .
$\frac{13}{2 x - 1}$ $=$ $- \frac{1}{5} + \frac{16}{15}$
$\frac{13}{2 x - 1}$ $=$ $- \frac{3}{15} + \frac{16}{15}$
$\frac{13}{2 x - 1}$ $=$ $\frac{13}{15}$
↓
invert the fractions
$\frac{2 x - 1}{13}$ $=$ $\frac{15}{13}$ $\cdot 13$
$2 x - 1$ $=$ $15$ $+ 1$
$2 x$ $=$ $16$ $: 2$
$x$ $=$ $8$
Now, plug $x$ into one of the above equations, e.g., $(I)$ , and solve for $y$ .
$(I) \frac{3}{2 x - 1} - \frac{8}{3 y + 2}$ $=$ $- \frac{1}{5}$
↓
plug in $x = 8$
$\frac{3}{2 \cdot 8 - 1} - \frac{8}{3 y + 2}$ $=$ $- \frac{1}{5}$
$\frac{3}{15} - \frac{8}{3 y + 2}$ $=$ $- \frac{1}{5}$
$\frac{1}{5} - \frac{8}{3 y + 2}$ $=$ $- \frac{1}{5}$ $- \frac{1}{5}$
$- \frac{8}{3 y + 2}$ $=$ $- \frac{2}{5}$
↓
invert the fractions
$- \frac{3 y + 2}{8}$ $=$ $- \frac{5}{2}$ $\cdot (- 8)$
$3 y + 2$ $=$ $\frac{5 \cdot 8}{2}$
$3 y + 2$ $=$ $20$ $- 2$
$3 y$ $=$ $18$ $: 3$
$y$ $=$ $6$
Write down the solution set, using first the solution for $x$ , then for $y$ .
$L = {(8 | 6)}$
Try to solve these non-linear equations using the methods you already know for linear systems of equations.

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$(II) \frac{2}{x} - \frac{4}{y}$	$=$	$- \frac{1}{6}$	$\cdot 2$
$({II}^{'}) \frac{4}{x} - \frac{8}{y}$	$=$	$- \frac{2}{6}$

$\frac{8}{x}$	$=$	$\frac{5}{3} - \frac{2}{6}$
$\frac{8}{x}$	$=$	$\frac{5}{3} - \frac{1}{3}$
$\frac{8}{x}$	$=$	$\frac{4}{3}$
	↓	invert the fractions
$\frac{x}{8}$	$=$	$\frac{3}{4}$	$\cdot 8$
$x$	$=$	$\frac{3 \cdot 8}{4}$
$x$	$=$	$6$

$(I) \frac{4}{x} + \frac{8}{y}$	$=$	$\frac{5}{3}$
	↓	plug in $x = 6$
$\frac{4}{6} + \frac{8}{y}$	$=$	$\frac{5}{3}$	$- \frac{4}{6}$
$\frac{8}{y}$	$=$	$\frac{5}{3} - \frac{4}{6}$
$\frac{8}{y}$	$=$	$\frac{5}{3} - \frac{2}{3}$
$\frac{8}{y}$	$=$	$\frac{3}{3}$
$\frac{8}{y}$	$=$	$1$	$\cdot y$
$y$	$=$	$8$

$(II) \frac{4}{y} + \frac{5}{2}$	$=$	$\frac{9}{x}$	$- \frac{9}{x} - \frac{5}{2}$
$({II}^{'}) - \frac{9}{x} + \frac{4}{y}$	$=$	$- \frac{5}{2}$

$({II}^{'}) - \frac{9}{x} + \frac{4}{y}$	$=$	$- \frac{5}{2}$	$\cdot 3$
$({II}^{''}) - \frac{27}{x} + \frac{12}{y}$	$=$	$- \frac{15}{2}$

$- \frac{20}{x}$	$=$	$- \frac{40}{6}$
$- \frac{x}{20}$	$=$	$- \frac{6}{40}$	$\cdot (- 20)$
$x$	$=$	$\frac{6 \cdot 20}{2 \cdot 20}$
	↓	shorten
$x$	$=$	$\frac{6}{2}$
$x$	$=$	$3$

$(II) \frac{4}{y} + \frac{5}{2}$	$=$	$\frac{9}{x}$
	↓	plug in $x = 3$
$\frac{4}{y} + \frac{5}{2}$	$=$	$\frac{9}{3}$	$- \frac{5}{2}$
$\frac{4}{y}$	$=$	$\frac{9}{3} - \frac{5}{2}$
$\frac{4}{y}$	$=$	$\frac{18}{6} - \frac{15}{6}$
$\frac{4}{y}$	$=$	$\frac{3}{6}$
$\frac{4}{y}$	$=$	$\frac{1}{2}$
	↓	invert the fraction
$\frac{y}{4}$	$=$	$\frac{2}{1}$	$\cdot 4$
$y$	$=$	$8$

$(I) \frac{4}{3 x + 1}$	$=$	$\frac{2}{3 y - 13}$	$\cdot (3 x + 1) \cdot (3 y - 13)$
	↓	multiply by the denominators
$4 \cdot (3 y - 13)$	$=$	$2 \cdot (3 x + 1)$
$(I^{'}) 12 y - 52$	$=$	$6 x + 2$

$(II) \frac{2}{5 x - 10}$	$=$	$\frac{4}{7 y - 6}$	$\cdot (5 x - 10) \cdot (7 y - 6)$
	↓	multiply by the denominators
$2 \cdot (7 y - 6)$	$=$	$4 \cdot (5 x - 10)$
$({II}^{'}) 14 y - 12$	$=$	$20 x - 40$

$(I^{'}) 12 y - 52$	$=$	$6 x + 2$	$- 2$
$12 y - 54$	$=$	$6 x$	$: 6$
$(I^{''}) 2 y - 9$	$=$	$x$

$(II) \frac{5}{2 x - 1} + \frac{4}{3 y + 2}$	$=$	$\frac{8}{15}$	$\cdot 2$
$({II}^{'}) \frac{10}{2 x - 1} + \frac{8}{3 y + 2}$	$=$	$\frac{16}{15}$

$(I) \frac{3}{2 x - 1} - \frac{8}{3 y + 2}$	$=$	$- \frac{1}{5}$
	↓	plug in $x = 8$
$\frac{3}{2 \cdot 8 - 1} - \frac{8}{3 y + 2}$	$=$	$- \frac{1}{5}$
$\frac{3}{15} - \frac{8}{3 y + 2}$	$=$	$- \frac{1}{5}$
$\frac{1}{5} - \frac{8}{3 y + 2}$	$=$	$- \frac{1}{5}$	$- \frac{1}{5}$
$- \frac{8}{3 y + 2}$	$=$	$- \frac{2}{5}$
	↓	invert the fractions
$- \frac{3 y + 2}{8}$	$=$	$- \frac{5}{2}$	$\cdot (- 8)$
$3 y + 2$	$=$	$\frac{5 \cdot 8}{2}$
$3 y + 2$	$=$	$20$	$- 2$
$3 y$	$=$	$18$	$: 3$
$y$	$=$	$6$

$({II}^{'}) 14 y - 12$	$=$	$20 x - 40$
	↓	plug in $x = 2 y - 9$
$14 y - 12$	$=$	$20 \cdot (2 y - 9) - 40$
	↓	Solve for $y$
$14 y - 12$	$=$	$40 y - 180 - 40$
$14 y - 12$	$=$	$40 y - 220$	$- 14 y + 220$
$208$	$=$	$26 y$	$: 26$
$y$	$=$	$8$

$(I^{'}) x$	$=$	$2 y - 9$
	↓	plug in $y = 8$
	$=$	$2 \cdot 8 - 9$
	$=$	$16 - 9$
	$=$	$7$

$(I) \frac{4}{3 x + 1}$	$=$	$\frac{2}{3 y - 13}$
	↓	plug in $x = 7$ and $y = 8$
$\frac{4}{3 \cdot 7 + 1}$	$=$	$\frac{2}{3 \cdot 8 - 13}$
$\frac{4}{21 + 1}$	$=$	$\frac{2}{24 - 13}$
$\frac{4}{22}$	$=$	$\frac{2}{11}$
$\frac{2}{11}$	$=$	$\frac{2}{11} ✓$

$\frac{13}{2 x - 1}$	$=$	$- \frac{1}{5} + \frac{16}{15}$
$\frac{13}{2 x - 1}$	$=$	$- \frac{3}{15} + \frac{16}{15}$
$\frac{13}{2 x - 1}$	$=$	$\frac{13}{15}$
	↓	invert the fractions
$\frac{2 x - 1}{13}$	$=$	$\frac{15}{13}$	$\cdot 13$
$2 x - 1$	$=$	$15$	$+ 1$
$2 x$	$=$	$16$	$: 2$
$x$	$=$	$8$

Exercises: systems with two variables

Solution by addition method

Solution by the addition method

Solution by the substitution method

Additionally: Check your solution

Solution using the addition method