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How many solutions does the system have? $\newline$ $3x+y=8$ $\newline$ $2x+2y=8$

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Find limits involving trigonometric functions

Full solution

Q. How many solutions does the system have?

\newline

3x+y=8

\newline

2x+2y=8

Write Equations: Write down the system of equations. $\newline$ $3x + y = 8$ $\newline$ $2x + 2y = 8$
Elimination Method: Try to solve the system using the elimination method. To do this, we can multiply the first equation by $2$ to match the coefficients of $y$ in the second equation. $\newline$ $(3x + y = 8) \times 2$ gives us $6x + 2y = 16$
Subtract Equations: Now we have the system: $\newline$ $6x + 2y = 16$ $\newline$ $2x + 2y = 8$ $\newline$ Subtract the second equation from the first to eliminate $y$ . $\newline$ $(6x + 2y) - (2x + 2y) = 16 - 8$ $\newline$ This simplifies to $4x = 8$
Solve for x: Solve for x. $\newline$ $4x = 8$ $\newline$ $x = \frac{8}{4}$ $\newline$ $x = 2$
Substitute $x$ for $y$ : Substitute $x$ back into one of the original equations to solve for $y$ . Using the first equation: $3x + y = 8$ $3(2) + y = 8$ $6 + y = 8$ $y = 8 - 6$ $y = 2$
Check Solution: Check the solution $(x=2, y=2)$ in the second equation to ensure it is correct. $2x + 2y = 8$ $2(2) + 2(2) = 8$ $4 + 4 = 8$ $8 = 8$ The solution satisfies the second equation as well.
Final Conclusion: Since both equations are satisfied by the solution $(x=2, y=2)$ , the system has exactly one solution.

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