The system of equations provides a unique solution(one solution) at (-0.6, 1.6).
As we already have the value of y from equation 1, substitute its value in equation 2,
[tex]6x-4y=-10[/tex]
[tex]6x-4(\dfrac{2}{3}x+2)=-10[/tex]
[tex]6x-\dfrac{8}{3}x-8=-10[/tex]
[tex]\dfrac{6x}{1}-\dfrac{8}{3}x=-10+8[/tex]
[tex]\dfrac{6x\times 3}{1\times 3}-\dfrac{8}{3}x=-2[/tex]
[tex]\dfrac{18x-8x}{3}=-2[/tex]
[tex]10x=-6\\[/tex]
[tex]x=\dfrac{-6}{10} \\\\x=-0.6[/tex]
[tex]y=\dfrac{2}{3}x+2[/tex]
substitute the value of x in equation 1,
[tex]y=\dfrac{2}{3}(\dfrac{-6}{10})+2\\\\y = \dfrac{-2}{5}+2\\\\y= \dfrac{-2+10}{5}\\\\y = \dfrac{8}{5}\\\\y=1.6[/tex]
As we can see the system of equations provides a unique solution(one solution) at (-0.6, 1.6).
Learn more about the system of equations:
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