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The straight line $\,2x\,+\,3y\,=\,5\,$ can be allotted in a graph by arranging it in a form of

\[\] \[y=mx+c\]. Where m is considered slope and c is considered y intercept.\[\]

After arranging $\,2x\,+\,3y\,=\,5\,$to \[y=-\dfrac{2}{3}x+\dfrac{5}{3}\] we get slope (\[m\]) as \[-\dfrac{2}{3}\]and y-intercept as \[-\dfrac{2}{3}\].

Thus, from the slope we conclude that the line is going downward (-ve slope means downward) , that is , it is decreasing. And the intercept shows that the line intercept at the y-axis at (0, 1.666667 ). [\[\dfrac{5}{3}\]=1.66667]

Now, to check whether the points (-3,4) and (7,-3) are solutions of this given line we need to put and of the points’ \[x\text{ }\!\!'\!\!\text{ s or }y's\]value in the equation.

Putting the value of \[x=-3\] from (-3,4) in the equation we get,

\[\begin{align}

& y=-\dfrac{2}{3}\times (-3)+\dfrac{5}{3} \\

& y=2+\dfrac{5}{3};\,\,\,\,\,\,\,\,\,\,\,y=\dfrac{6+5}{3}; \\

& \therefore y=\dfrac{11}{3}\text{ which is not equal to y=4 from the points given to us}\text{.} \\

& \therefore \text{(-3,4) is not a solution for this line}\text{.} \\

\end{align}\]

We again put,

\[\begin{align}

& x=7\text{ from }(7,-3) \\

& y=-\dfrac{2}{3}\times (7)+\dfrac{5}{3} \\

& \therefore y=-\dfrac{14}{3}+\dfrac{5}{3}\,\,\,\,\,\,\,\,\,\,y=\dfrac{-14+5}{3} \\

& \therefore y=\dfrac{-9}{3}=-3\text{ which is equal to }y=-3\text{ from the point (7,-3)} \\

& \therefore \text{ (7,-3) is a solution for this line}\text{.} \\

\end{align}\]