Exercises: Intersection of two lines

Calculate the intersection of the pairs of lines.

$y = 3 x + 4$ and $y = - 2 x + 14$

$y = 6 x - 3$ and $y = 7 x - 11$

$y = 8 x + 3$ and $y = - 4 x + 6$

$y = \frac{1}{6} x - 4$ and $y = \frac{1}{3} x - 10$

$y = \frac{1}{2} x + \frac{3}{2}$ and $y = \frac{1}{2}$

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$y$	$=$	$3 x + 4$
$y$	$=$	$- 2 x + 14$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$3 x + 4$	$=$	$- 2 x + 14$	$+ 2 x; - 4$
$5 x$	$=$	$10$	$: 5$
$x$	$=$	$2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$y$	$=$	$3 \cdot 2 + 4$
	↓	First multiply, then divide.
$y$	$=$	$10$
	↓	$\Rightarrow$ Intersection point: $S (2 \| 10)$

$y$	$=$	$6 x - 3$
$y$	$=$	$7 x - 11$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$6 x - 3$	$=$	$7 x - 11$	$+ 3; - 7 x$
$- x$	$=$	$- 8$	$: (- 1)$
$x$	$=$	$8$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$y$	$=$	$6 \cdot 8 - 3$
	↓	First multiply, then divide.
$y$	$=$	$45$
	↓	$\Rightarrow$ Intersection point: $S (8 \| 45)$

$y$	$=$	$8 x + 3$
$y$	$=$	$- 4 x + 6$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$8 x + 3$	$=$	$- 4 x + 6$	$- 3; + 4 x$
$12 x$	$=$	$3$	$: 12$
$x$	$=$	$\frac{3}{12} = \frac{1}{4}$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$y$	$=$	$8 \cdot \frac{1}{4} + 3$
	↓	First multiply, then divide.
$y$	$=$	$5$
	↓	$\Rightarrow$ Intersection point: $S (\frac{1}{4} \| 5)$

$y$	$=$	$\frac{1}{6} x - 4$
$y$	$=$	$\frac{1}{3} x - 10$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\frac{1}{6} x - 4$	$=$	$\frac{1}{3} x - 10$	$- \frac{1}{3} x; + 4$
$- \frac{1}{6 x}$	$=$	$- 6$	$: (- \frac{1}{6})$
$x$	$=$	$36$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$y$	$=$	$\frac{1}{3} \cdot 36 - 10$
	↓	First multiply, then divide.
$y$	$=$	$2$
	↓	$\Rightarrow$ Intersection point: $S (36 \| 2)$

$y$	$=$	$\frac{1}{2} x + \frac{3}{2}$
$y$	$=$	$\frac{1}{2}$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\frac{1}{2} x + \frac{3}{2}$	$=$	$\frac{1}{2}$	$- \frac{3}{2}$
$\frac{1}{2} x$	$=$	$- 1$	$\cdot 2$
$x$	$=$	$- 2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$y$	$=$	$\frac{1}{2}$
	↓	$\Rightarrow$ Intersection point: $S (- 2 \| 0.5)$