14Example: Elimination by equation coefficients

Consider the linear system $\def\arraystretch{2} \begin{array} {rrl} \\\mathrm{I}) &-3y + 15x &= 3 \\\; \\\mathrm{II}) &-\dfrac{1}{2}x - \dfrac{5}{2} &= \dfrac{5}{2}y\end{array}$

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Calculate the coordinates of the intersection point of the two lines using elimination by equationg coefficients! Berechne die Koordinaten des Schnittpunktes der zwei Geraden, indem du dazu das Gleichsetzungsverfahren benutzt!

Solution

To apply the elimination by equationg coefficients, you must solve the two equations either for $x$ or for $y$. Since a line equation has the form $y = mx + t$, it is easier to solve the equations for $y$.

Keep in mind how to transform equations.

Solve for one variable

$\def\arraystretch{2} \begin{array} {rrl} \mathrm{I}) &-3y + 15x &= 3 & \mid-15x\\\mathrm{II}) &\dfrac{5}{2}y &= -\dfrac{1}{2}x - \dfrac{5}{2}\end{array}$

$\def\arraystretch{2} \begin{array} {rrll} \mathrm{I})' &-3y &= - 15x + 3 &\mid :(-3)\\\mathrm{II}) &\dfrac{5}{2}y &= -\dfrac{1}{2}x - \dfrac{5}{2}& \mid\cdot \dfrac{2}{5}\end{array}$

$\def\arraystretch{2} \begin{array} {rrl} \mathrm{I})' &y &= 5x -1 \\\mathrm{II})' &y &= -\dfrac{1}{5}x - 1\end{array}$