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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$\displaystyle y$	$=$	$\displaystyle 3x+4$
$\displaystyle y$	$=$	$\displaystyle -2x+14$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle 3x+4$	$=$	$\displaystyle -2x+14$	$\displaystyle +2x;\ -4$
$\displaystyle 5x$	$=$	$\displaystyle 10$	$\displaystyle :5$
$\displaystyle x$	$=$	$\displaystyle 2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle 3\cdot2+4$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 10$
	↓	$\;\;\Rightarrow\;\;$ Intersection point: $S (2 ∣ 10)$

$\displaystyle y$	$=$	$\displaystyle 6x-3$
$\displaystyle y$	$=$	$\displaystyle 7x-11$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle 6x-3$	$=$	$\displaystyle 7x-11$	$\displaystyle +3;\ -7x\$
$\displaystyle -x$	$=$	$\displaystyle -8$	$\displaystyle :\left(-1\right)$
$\displaystyle x$	$=$	$\displaystyle 8$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle 6\cdot8-3$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 45$
	↓	$\Rightarrow$ Intersection point: $S (8 ∣ 45)$

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

$\displaystyle y$	$=$	$\displaystyle \frac{1}{6}x-4$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{3}x-10$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle \frac{1}{6}x-4$	$=$	$\displaystyle \frac{1}{3}x-10$	$\displaystyle -\frac{1}{3}x;+4$
$\displaystyle -\frac{1}{6x}$	$=$	$\displaystyle -6$	$\displaystyle :\left(-\frac{1}{6}\right)$
$\displaystyle x$	$=$	$\displaystyle 36$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle \frac{1}{3}\cdot36-10$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 2$
	↓	$\Rightarrow$ Intersection point: $S (36 ∣ 2)$

$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}x+\frac{3}{2}$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle \frac{1}{2}x+\frac{3}{2}$	$=$	$\displaystyle \frac{1}{2}$	$\displaystyle -\frac{3}{2}$
$\displaystyle \frac{1}{2}x$	$=$	$\displaystyle -1$	$\displaystyle \cdot2$
$\displaystyle x$	$=$	$\displaystyle -2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}$
	↓	$\Rightarrow$ Intersection point: $S (- 2 ∣ 0.5)$

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