We should determine such values (x,y) that satisfy both equations:
"\\begin{cases}\n3x-6y=10, \\\\ 9x+15y=-14\n\\end{cases}"
Let us multiply the first equation by 3 and subtract the second equation from the first:
"\\begin{cases}\n9x-18y=30, \\\\ 9x+15y=-14\n\\end{cases}" , "-33y = 44 \\Rightarrow y = -\\dfrac43\\,."
Next, we substitute the value of y into the first equation and obtain the value of x:
"x = \\dfrac19(30 + 18y) = \\dfrac19(30- 24) = \\dfrac23."
Therefore, "(x,y) = \\left( \\dfrac23, -\\dfrac43\\right)."
Comments
Leave a comment