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Easy Level Solve the system:$\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}$
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Solution:
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The coefficients of y in the two equations are opposites (1 and -1). Add the two equations to eliminate y:$(x + y) + (x - y) = 5 + 1$ Simplify: $2x = 6$, so $x = 3$.
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Substitute $x = 3$ into the first equation: $3 + y = 5$, so $y = 2$. The solution to the system is $\begin{cases} x = 3 \\ y = 2 \end{cases}$.
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Medium Level Solve the system:$\begin{cases} 2x + 3y = 8 \\ 4x - y = 7 \end{cases}$
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Solution:
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Observe the coefficients: the coefficients of x are 2 and 4 (in a multiple relationship). Multiply the first equation by 2 to make the coefficients of x equal: First equation × 2: $4x + 6y = 16$ (denoted as Equation ③)
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Subtract the second equation from Equation ③ to eliminate x:$(4x + 6y) - (4x - y) = 16 - 7$ Simplify: $7y = 9$, so $y = \frac{9}{7}$.
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Substitute $y = \frac{9}{7}$ into the second equation: $4x - \frac{9}{7} = 7$, so $x = \frac{29}{14}$. The solution to the system is $\begin{cases} x = \frac{29}{14} \\ y = \frac{9}{7} \end{cases}$.
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Hard Level
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Solve the system:$\begin{cases} 3x - 2y = 10 \\ 5x + 3y = 4 \end{cases}$
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Solution: To eliminate y, multiply the equations by constants to make the coefficients of y opposites: First equation × 3: $9x - 6y = 30$ (denoted as Equation ③) Second equation × 2: $10x + 6y = 8$ (denoted as Equation ④)
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Add Equation ③ and Equation ④ to eliminate y:$(9x - 6y) + (10x + 6y) = 30 + 8$ Simplify: $19x = 38$, so $x = 2$.
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Substitute $x = 2$ into the first equation: $3×2 - 2y = 10$, so $y = -2$. The solution to the system is $\begin{cases} x = 2 \\ y = -2 \end{cases}$.