Question

Help please!!! Systems Of Equations

Answer 1

$6x+3y=-3\qquad\text{subtract 6x from both sides}\\3y=-6x-3\qquad\text{divide both sides by 3}\\y=-2x-1\\\\\left\{\begin{array}{ccc}y=-2x+5\\y=-2x-1\end{array}\right\\\\(1)\qquad y=-2x+5\\\\for\ x=0\to y=-2(0)+5=0+5=5\to(0,\ 5)\\for\ x=2\to y=-2(2)+5=-4+5=1\to(2,\ 1)\\\\(2)\qquad y=-2x-1\\\\for\ x=0\to y=-2(0)-1=0-1=-1\to(0,\ -1)\\for\ x=-2\to y=-2(-2)-1=4-1=3\to(-2,\ 3)\\\\Answer:\ NO\ SOLUTION\ \text{(An inconsistent system of equations)}$

$4x-6y=24\qquad\text{subtract 4x from both sides}\\-6y=-4x+24\qquad\text{divide both sides by (-4)}\\y=\dfrac{2}{3}x-4\\\\\left\{\begin{array}{ccc}y=-\dfrac{1}{3}x+2\\\\y=\dfrac{2}{3}x-4\end{array}\right\\\\(1)\qquad y=-\dfrac{1}{3}x+2\\\\for\ x=0\to y=-\dfrac{1}{3}(0)+2=0+2=2\to(0,\ 2)\\\\for\ x=3\to y=-\dfrac{1}{3}(3)+3=-1+2=1\to(3,\ 1)\\\\(2)\qquad y=\dfrac{2}{3}x-4\\\\for\ x=0\to y=\dfrac{2}{3}(0)-4=0-4=-4\to(0,\ -4)\\\\for\ x=3\to y=\dfrac{2}{3}(3)-4=2-4=-\to(3,\ -2)\\\\Answer:\ (6,\ 0)\to x=6\ and\ y=0$