One of the solutions of the linear equation 3 | vedclass

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Question

(A) To find a solution for the linear equation $3x + 5y = 30$,we can substitute the given options into the equation.For option $A$ $(10, 0)$,substitut

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$(10, 0)$

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(A) To find a solution for the linear equation $3x + 5y = 30$,we can substitute the given options into the equation.For option $A$ $(10, 0)$,substitute $x = 10$ and $y = 0$:$3(10) + 5(0) = 30 + 0 = 30$.Since the left-hand side equals the right-hand side,$(10, 0)$ is a solution.For option $B$ $(0, 10)$,substitute $x = 0$ and $y = 10$:$3(0) + 5(10) = 0 + 50 = 50 
eq 30$.For option $C$ $(3, 5)$,substitute $x = 3$ and $y = 5$:$3(3) + 5(5) = 9 + 25 = 34 
eq 30$.For option $D$ $(0, 0)$,substitute $x = 0$ and $y = 0$:$3(0) + 5(0) = 0 
eq 30$.Therefore,the correct solution is $(10, 0)$.

One of the solutions of the linear equation $3x + 5y = 30$ is $\ldots \ldots . .$

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