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How to Verify if Ordered Pairs Are Solutions to a Linear System 2x+y=1 and 3x+2y=5
Mathematics
Grade 8 of Junior High School
Question Content
Verify if the ordered pairs (6, 1) and (18, 6) are solutions to the system of linear equations: $\\begin{cases}2x + y = 1 \\\\ 3x + 2y = 5\\end{cases}$
Correct Answer
Neither (6, 1) nor (18, 6) is a solution to the given system of linear equations; the actual solution to the system is (-3, 7)
Detailed Solution Steps
1
Step 1: Verify the ordered pair (6, 1). Substitute x=6 and y=1 into the first equation: $2(6) + 1 = 12 + 1 = 13$, which is not equal to 1. Since it fails the first equation, (6,1) is not a solution.
2
Step 2: Verify the ordered pair (18, 6). Substitute x=18 and y=6 into the first equation: $2(18) + 6 = 36 + 6 = 42$, which is not equal to 1. Since it fails the first equation, (18,6) is not a solution.
3
Step 3 (optional: solve the system to find the actual solution): Multiply the first equation by 2: $4x + 2y = 2$. Subtract the second equation from this new equation: $(4x + 2y) - (3x + 2y) = 2 - 5$, which simplifies to $x = -3$. Substitute x=-3 into the first equation: $2(-3) + y = 1$, so $-6 + y = 1$, and $y = 7$. The actual solution is (-3,7).
Knowledge Points Involved
1
Solution to a System of Linear Equations
An ordered pair (x,y) is a solution to a system of linear equations if and only if it satisfies all equations in the system simultaneously. This means substituting the x and y values into each equation results in a true mathematical statement.
2
Substitution Method for Verifying Solutions
To check if an ordered pair is a solution, replace the variables x and y in each equation with the corresponding values from the pair, then simplify both sides of the equation to check for equality. If any equation fails this check, the pair is not a solution.
3
Elimination Method for Solving Linear Systems
The elimination method involves multiplying one or both equations by a constant to make the coefficients of one variable opposites, then adding or subtracting the equations to eliminate that variable. This allows solving for the remaining variable, which can then be substituted back to find the other variable's value. It is used to find the unique solution (if it exists) for a system of linear equations.
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