Exercises: systems with two variables

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

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II})&\frac 2x-\frac 4y&=&-\frac16\end{array}$

where $x,y \neq 0$

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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: $\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II})&\frac 2x-\frac 4y&=&-\frac16\end{array}$

The equation $(\text{II})$ contains a term $-\frac{4}{y}$ and equation $(\text I)$ is a multiple of $-\frac{4}{y}$ , namely $\frac{8}{y}$ . So you can multiply $(\text{II})$ by $2$ and add both equations to eliminate $\frac{8}{y}$ .

$\displaystyle \left(\text{II}\right)\ \ \frac{2}{x}-\frac{4}{y}$	$=$	$\displaystyle -\frac{1}{6}$	$\displaystyle \cdot2$
$\displaystyle \left(\text{II}'\right)\ \ \frac{4}{x}-\frac{8}{y}$	$=$	$\displaystyle -\frac{2}{6}$

So the new system of equations is:

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\(\text{II}')&\frac 4x-\frac 8y&=&-\frac26\end{array}$

$(\text I) + (\text{II}')$ :

$\def\arraystretch{1.25} \begin{array}{rcll}(\text{I})&\frac 4x+\frac 8y&=&\frac53\\+\;\;(\text{II}')&\frac 4x-\frac 8y&=&-\frac26\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{8}{x}\;\;\;\;\;\;\;\;\;=\;\frac{5}3-\frac{2}{6}\;\;\;}$

Now solve the new equation for $x$ .

$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{5}{3}-\frac{2}{6}$
$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{5}{3}-\frac{1}{3}$
$\displaystyle \frac{8}{x}$	$=$	$\displaystyle \frac{4}{3}$
	↓	invert the fractions
$\displaystyle \frac{x}{8}$	$=$	$\displaystyle \frac{3}{4}$	$\displaystyle \cdot8$
$\displaystyle x$	$=$	$\displaystyle \frac{3\cdot8}{4}$
$\displaystyle x$	$=$	$\displaystyle 6$

Now, plug $x$ into one of the above equations, e.g. $(\text{I})$ , and solve for $y$ .

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

Write down the solution set. $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(6\;\left|\;8\right.\right)\right\}$ .

Do you have a question?

Write it here 👇

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}\right)&\frac{4}{y}+\frac{5}{2}&=&\frac{9}{x}\end{array}$

where $x,y \neq 0$

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.

Solution by the addition method

Using the addition method you get to the following solution:

Given is: $\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}\right)&\frac{4}{y}+\frac{5}{2}&=&\frac{9}{x}\end{array}$

Transform $(\text{II})$ such that on one side you have only variables and on the other side only numbers.

$\displaystyle (\text{II})\;\;\;\;\;\;\;\;\;\frac4y+\frac52$	$=$	$\displaystyle \frac{9}{x}\ \ \ \ \ \ \$	$\displaystyle -\frac{9}{x}-\frac{5}{2}$
$\displaystyle (\text{II}')\;\;\;\;-\frac{9}{x}+\frac{4}{y}$	$=$	$\displaystyle -\frac{5}{2}$

The new system of equations is now:

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\\left(\text{II}'\right)&-\frac{9}{x}+\frac{4}{y}&=&-\frac{5}{2}\end{array}$

Now, equation $(\text{II}')$ contains the term $\frac{4}{y}$ and equation $(\text I)$ is a multiple of $\frac{4}{y}$ , namely $-\frac{12}{y}$ . So you multiply $(\text{II}')$ by $3$ and add both equations in order to eliminate $\frac{12}{y}$ .

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

$(\text I) +(\text{II}'')$ :

$\left(I\right)\frac7x-\frac{12}y=\frac56$

$\left(II\right)-\frac{27}{x}+\frac{12}{y}=-\frac{15}{2}$

$\def\arraystretch{1.25} \begin{array}{rrcr}(\text I)&\frac{7}{x}-\frac{12}{y}&=&\frac{5}{6}\\+\left(\text{II}''\right)&-\frac{27}{x}+\frac{12}{y}&=&-\frac{15}{2}\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-\frac{20}x\;=\;-\frac{40}6\;\;\;}$

Now, you solve $-\frac{20}x\;=\;-\frac{40}6$ , by inverting the fraction.

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

Now, plug $x$ into one of the above equations, for instance $(\text{II})$ , and solve for $y$ .

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

Write down the solution set. $L=\left\{\left(x\;\left|\;y\right.\right)\right\}=\left\{\left(3\;\left|\;8\right.\right)\right\}$

Do you have a question?

Write it here 👇

$\def\arraystretch{1.25} \begin{array}{l}(\text I)\;\;\frac{4}{3x+1}=\frac{2}{3y-13}\\(\text{II})\;\frac 2{5x-10}=\frac{4}{7y-6}\end{array}$

where $x\notin\left\{-\frac{1}{3};2\right\}$ and $y\notin\left\{\frac{13}{3};\frac{6}{7}\right\}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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: $\def\arraystretch{1.25} \begin{array}{l}(\text{I})\;\;\frac{4}{3x+1}=\frac{2}{3y-13}\\(\text{II})\;\frac 2{5x-10}=\frac{4}{7y-6}\end{array}$

Transform $(\text{I})$ and $(\text{II})$ , such that both equations no longer contain a fraction.

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

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

Thus, the linear system of equations is:

$\def\arraystretch{1.25} \begin{array}{l}(\text{I}')\;\;12y-52=6x+2\\(\text{II}')\;14y-12=20x-40\end{array}$

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: $\def\arraystretch{1.25} \begin{array}{l}(\text I')\;\;12y-52=6x+2\\(\text{II}')\;14y-12=20x-40\end{array}$

Transform $(\text{I}')$ , such that $x$ appears only one one side.

$\displaystyle \left(\text{I}'\right)\ \ \ \ \ 12y-52$	$=$	$\displaystyle 6x+2$	$\displaystyle -2$
$\displaystyle 12y-54$	$=$	$\displaystyle 6x$	$\displaystyle :6$
$\displaystyle \left(\text{I}''\right)\ \ \ \ \ \ \ \ 2y-9$	$=$	$\displaystyle x$

Plug $x = 2 y - 9$ into $(\text{II}')$ .

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

Plug $y = 8$ into $(\text I')$ , in order to find $x$ .

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

Write down the solution set. $L=\left\{\left(x\left|y\right.\right)\right\}=\left\{\left(7\left|8\right.\right)\right\}$

Additionally: Check your solution

Plug $x = 7$ and $y = 8$ into equations $(\text I)$ and $(\text{II})$ and check if the equation is satisfied.

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

$\displaystyle (\text{II})\;\frac 2{5x-10}$	$=$	$\displaystyle \frac{4}{7y-6}$
	↓	plug in $x = 7$ and $y = 8$
$\displaystyle \frac{2}{5\cdot 7-10}$	$=$	$\displaystyle \frac{4}{7\cdot 8-6}$
$\displaystyle \frac{2}{35-10}$	$=$	$\displaystyle \frac{4}{56-6}$
$\displaystyle \frac{2}{25}$	$=$	$\displaystyle \frac{4}{50}$
$\displaystyle \frac{2}{25}$	$=$	$\displaystyle \frac{2}{25}\ \ \ \ \ \ \ \checkmark$

Both equations are satisfied, so our solution is correct.

Do you have a question?

Write it here 👇

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac{3}{2x-1}-\frac{8}{3y+2}&=&-\frac{1}{5}\\\left(\text{II}\right)&\frac5{2x-1}+\frac4{3y+2}&=&\frac8{15}\end{array}$

where $x\ne\frac{1}{2}$ and $y \neq -\frac 23$

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.

Solution using the addition method

Given is: $\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\\left(\text{II}\right)&\frac5{2x-1}+\frac4{3y+2}&=&\frac{8}{15}\end{array}$

The equation $(\text{II})$ contains the term $\frac{4}{3y+2}$ and $(\text I)$ contains a multiple of $\frac{4}{3y+2}$ , namely $-\frac{8}{3y+2}$ . So you multiply $(\text{II})$ by $2$ and add both equations in order to eliminate $\frac{8}{3y+2}$ .

$\displaystyle \left(\text{II}\right)\ \ \frac{5}{2x-1}+\frac{4}{3y+2}$	$=$	$\displaystyle \frac{8}{15}\ \ \ \ \$	$\displaystyle \cdot2$
$\displaystyle \left(\text{II}'\right)\ \ \frac{10}{2x-1}+\frac{8}{3y+2}$	$=$	$\displaystyle \frac{16}{15}$

So the new system of equations is:

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\\left(\text{II}'\right)&\frac{10}{2x-1}+\frac8{3y+2}&=&\frac{16}{15}\end{array}$

$(\text I)+(\text{II}')$ :

$\def\arraystretch{1.25} \begin{array}{rrcl}(\text I)&\frac3{2x-1}-\frac8{3y+2}&=&-\frac{1}{5}\\+\left(\text{II}'\right)&\frac{10}{2x-1}+\frac8{3y+2}&=&\frac{16}{15}\end{array}$

$\overline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{13}{2x-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;-\frac{1}5+\frac{16}{15}\;\;\;}$

Now solve the equation resulting from the addition for $x$ .

$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle -\frac{1}{5}+\frac{16}{15}$
$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle -\frac{3}{15}+\frac{16}{15}$
$\displaystyle \frac{13}{2x-1}$	$=$	$\displaystyle \frac{13}{15}$
	↓	invert the fractions
$\displaystyle \frac{2x-1}{13}$	$=$	$\displaystyle \frac{15}{13}$	$\displaystyle \cdot13$
$\displaystyle 2x-1$	$=$	$\displaystyle 15$	$\displaystyle +1$
$\displaystyle 2x$	$=$	$\displaystyle 16$	$\displaystyle :2$
$\displaystyle x$	$=$	$\displaystyle 8$

Now, plug $x$ into one of the above equations, e.g., $(\text{I})$ , and solve for $y$ .

$\displaystyle \left(\text{I}\right)\ \ \frac{3}{2x-1}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
	↓	plug in $x = 8$
$\displaystyle \frac{3}{2\cdot8-1}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
$\displaystyle \frac{3}{15}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$
$\displaystyle \frac{1}{5}-\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{1}{5}$	$\displaystyle -\frac{1}{5}$
$\displaystyle -\frac{8}{3y+2}$	$=$	$\displaystyle -\frac{2}{5}$
	↓	invert the fractions
$\displaystyle -\frac{3y+2}{8}$	$=$	$\displaystyle -\frac{5}{2}$	$\displaystyle \cdot\left(-8\right)$
$\displaystyle 3y+2$	$=$	$\displaystyle \frac{5\cdot8}{2}$
$\displaystyle 3y+2$	$=$	$\displaystyle 20$	$\displaystyle -2$
$\displaystyle 3y$	$=$	$\displaystyle 18$	$\displaystyle :3$
$\displaystyle y$	$=$	$\displaystyle 6$

Write down the solution set, using first the solution for $x$ , then for $y$ .

$L=\left\{\left(8\;\left|\;6\right.\right)\right\}$

Do you have a question?

Write it here 👇

This content is licensed under
CC BY-SA 4.0 → Info