Find the solution of the system of equations. {:[3x-3y=0],[15 x+6y=42]:}

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Find the solution of the system of equations.  {:[3x-3y=0],[15 x+6y=42]:}

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Analyze Equations: Analyze the system of equations. We have the system: 3x - 3y = 0 15x + 6y = 42 We need to find the values of x and y that satisfy both equations simultaneously.Simplify First Equation: Simplify the first equation. The first equation can be simplified by dividing both sides by 3 to isolate x in terms of y. (3x - 3y) / 3 = 0 / 3 x - y = 0 x = ySubstitute and Solve: Substitute x with y in the second equation. Since x = y, we can replace x with y in the second equation to find the value of y. 15y + 6y = 42Combine and Find y: Combine like terms and solve for y. 15y + 6y = 42 21y = 42 y = 42 / 21 y = 2Substitute Back for x: Substitute y back into x = y to find the value of x. Since x = y and we found y = 2, then x = 2.Check Solution: Check the solution in both original equations. First equation: 3x - 3y = 0 3(2) - 3(2) = 0 6 - 6 = 0 0 = 0 (True)  Second equation: 15x + 6y = 42 15(2) + 6(2) = 42 30 + 12 = 42 42 = 42 (True)  Both checks are correct, so the solution (x, y) = (2, 2) satisfies both equations.

Find the solution of the system of equations. $\newline$ $\begin{aligned} 3 x-3 y & =0 \\ 15 x+6 y & =42 \end{aligned}$

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Find the solution of the system of equations.\newline3x−3yamp;=015x+6yamp;=42 \begin{aligned} 3 x-3 y &amp; =0 \\ 15 x+6 y &amp; =42 \end{aligned} 3x−3y15x+6y​amp;=0amp;=42​

Find the solution of the system of equations. $\newline$ $\begin{aligned} 3 x-3 y & =0 \\ 15 x+6 y & =42 \end{aligned}$