Aus Online Mathematik Brückenkurs 1

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The lines have a point of intersection at that point which simultaneously satisfies the equations of both lines {{Abgesetzte Formel||<math>2x+y-1=0\qquad\text{and}\qquad y-2x-2=0\,\textrm{.}</math>}} If we make ''y'' the subject of the second equation <math>y=2x+2</math> and substitute it into the first equation, we obtain an equation which only contains ''x'', {{Abgesetzte Formel||<math>2x+(2x+2)-1=0\quad\Leftrightarrow\quad 4x+1=0\,,</math>}} which gives that <math>x=-1/4\,</math>. Then, from the relation <math>y=2x+2</math>, we obtain <math>y = 2\cdot(-1/4)+2 = 3/2\,</math>. The point of intersection is <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math>. <center>[[Image:2_2_6_e.gif|center]]</center> We check for safety's sake that <math>\bigl(-\tfrac{1}{4},\tfrac{3}{2}\bigr)</math> really satisfies both equations: :*2''x''&nbsp;+&nbsp;''y''&nbsp;-&nbsp;1&nbsp;=&nbsp;0: <math>\quad\textrm{LHS} = 2\cdot\bigl(-\tfrac{1}{4}\bigr) + \tfrac{3}{2} - 1 = -\tfrac{1}{2} + \tfrac{3}{2} - \tfrac{2}{2} = 0 = \textrm{RHS.}</math> :*''y''&nbsp;-&nbsp;2''x''&nbsp;-&nbsp;2&nbsp;=&nbsp;0: <math>\quad\textrm{LHS} = \tfrac{3}{2}-2\cdot\bigl(-\tfrac{1}{4}\bigr)-2 = \tfrac{3}{2} + \tfrac{1}{2} - \tfrac{4}{2} = 0 = \textrm{RHS.}</math>

• 2.2.6e - Solution - The lines 2x + y -1 = 0 and y - 2x - 2 = 0 through the point (-1/4,3/2)
• Vorlage:Abgesetzte Formel

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