Equation 1: [tex]y = \frac{2}{3}x+2 [/tex]
Equation 2: [tex]6x-4y=-10[/tex]
Substitute Equation 1 into Equation 2
[tex]6x-4[ \frac{2}{3}x+2]=-10 [/tex]
[tex]6x- \frac{8}{3}x -8=-10[/tex]
[tex] \frac{10}{3}x-8=-10 [/tex]
[tex] \frac{10}{3}x=-10+8 [/tex]
[tex] \frac{10}{3}x=-2 [/tex]
[tex]10x=-6[/tex]
[tex]x=- \frac{6}{10} [/tex]
[tex]x=-0.6[/tex]
Substitute x = -0.6 into either Equation 1 or Equation 2 to work out 'y'
[tex]y= \frac{2}{3}(-0.6)+2 = 1.6 [/tex]
The linear system only have one solution (-0.6, 1.6)
Correct answer: B
