Exercises: systems with two variables

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

$\begin{array}{lrcll} (I) & 2 y & = & 2 x - 40 \\ (II) & 3 x & = & 10 - 2 y \end{array}$

For this task you need the following basic knowledge: System of linear equations
You can solve linear systems of equations using the substitution method, the addition method, or by equating coefficients.
The substitution method is best suited here.
Substitution method
Given is: $\begin{array}{l} (I) 2 y = 2 x - 40 \\ (II) 3 x = 10 - 2 y \end{array}$
Equation $(I)$ is solved for $2 y$ . This term also appears in equation $(II)$ . So plug equation $(I)$ into $(I I)$ .
$(II) 3 x$ $=$ $10 - 2 y$
↓
from eq $(I)$ , plug in $2 y = 2 x - 40$
$3 x$ $=$ $10 - (2 x - 40)$
↓
multiply out the bracket
$3 x$ $=$ $10 - (2 x - 40)$ $+ 2 x$
↓
solve for $x$
$5 x$ $=$ $50$ $: 5$
$x$ $=$ $10$
To find $y$ , plug the value of $x$ into $(I)$ .
$(I) 2 y$ $=$ $2 x - 40$
↓
plug in $x = 10$
$2 y$ $=$ $2 \cdot 10 - 40$
$2 y$ $=$ $20 - 40$
$2 y$ $=$ $- 20$ $: 2$
$y$ $=$ $- 10$
Write down the solution set.
$L = {(x | y)} = {(10 | - 10)}$
Solution by other methods
$\begin{array}{l} (I) \frac{1}{2} x - \frac{3}{5} y = 3 \\ (II) \frac{1}{4} x + y = 8 \end{array}$

For this task you need the following basic knowledge: System of linear equations
You can solve linear systems of equations using the substitution method, the addition method, or by equating coefficients.
The substitution method is best suited here.
Substitution method
Given is: $\begin{array}{l} (I) \frac{1}{2} x - \frac{3}{5} y = 3 \\ (II) \frac{1}{4} x + y = 8 \end{array}$
Rewrite $(I I)$ such that $y$ is on one side.
$(II) \frac{1}{4} x + y$ $=$ $8$ $- \frac{1}{4} x$
$(II’) y$ $=$ $8 - \frac{1}{4} x$
Plug $y = 8 - \frac{1}{4} x$ into $(I)$ and solve for $x$ .
$(I) \frac{1}{2} x - \frac{3}{5} y$ $=$ $3$
↓
plug in $y = 8 - \frac{1}{4} x$
$\frac{1}{2} x - \frac{3}{5} \cdot (8 - \frac{1}{4} x)$ $=$ $3$
$\frac{1}{2} x - \frac{24}{5} + \frac{3}{20} x$ $=$ $3$ $+ \frac{24}{5}$
$\frac{1}{2} x + \frac{3}{20} x$ $=$ $3 + \frac{24}{5}$
↓
compute the fractions
$\frac{10}{20} x + \frac{3}{20} x$ $=$ $3 + 4 \frac{4}{5}$
$\frac{13}{20} x$ $=$ $7 \frac{4}{5}$ $\cdot \frac{20}{13}$
$x$ $=$ $12$
plug $x = 12$ into $(II’)$ .
$(II’) y$ $=$ $8 - \frac{1}{4} x$
↓
plug in $x = 12$
$=$ $8 - \frac{1}{4} \cdot 12$
$=$ $8 - 3$
$=$ $5$
Write down the solution set, first the solution for $x$ , then for $y$ .
$L = {(12 | 5)}$
Solution by other methods

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$(II) 3 x$	$=$	$10 - 2 y$
	↓	from eq $(I)$ , plug in $2 y = 2 x - 40$
$3 x$	$=$	$10 - (2 x - 40)$
	↓	multiply out the bracket
$3 x$	$=$	$10 - (2 x - 40)$	$+ 2 x$
	↓	solve for $x$
$5 x$	$=$	$50$	$: 5$
$x$	$=$	$10$

$(I) 2 y$	$=$	$2 x - 40$
	↓	plug in $x = 10$
$2 y$	$=$	$2 \cdot 10 - 40$
$2 y$	$=$	$20 - 40$
$2 y$	$=$	$- 20$	$: 2$
$y$	$=$	$- 10$

$(II) \frac{1}{4} x + y$	$=$	$8$	$- \frac{1}{4} x$
$(II’) y$	$=$	$8 - \frac{1}{4} x$

$(I) \frac{1}{2} x - \frac{3}{5} y$	$=$	$3$
	↓	plug in $y = 8 - \frac{1}{4} x$
$\frac{1}{2} x - \frac{3}{5} \cdot (8 - \frac{1}{4} x)$	$=$	$3$
$\frac{1}{2} x - \frac{24}{5} + \frac{3}{20} x$	$=$	$3$	$+ \frac{24}{5}$
$\frac{1}{2} x + \frac{3}{20} x$	$=$	$3 + \frac{24}{5}$
	↓	compute the fractions
$\frac{10}{20} x + \frac{3}{20} x$	$=$	$3 + 4 \frac{4}{5}$
$\frac{13}{20} x$	$=$	$7 \frac{4}{5}$	$\cdot \frac{20}{13}$
$x$	$=$	$12$

$(II’) y$	$=$	$8 - \frac{1}{4} x$
	↓	plug in $x = 12$
	$=$	$8 - \frac{1}{4} \cdot 12$
	$=$	$8 - 3$
	$=$	$5$

Exercises: systems with two variables

Substitution method

Solution by other methods

Substitution method

Solution by other methods