Algebraic methods: Substitution method and Elimination method.

The second equation is (2) times the first, so both are the same line and have infinitely many solutions. Identify proportional equations.

Given \(x=5\), substitute it into \(2x-y=1\): \(2(5)-y=1\), so \(10-y=1\). Subtracting 10 from both sides gives \(-y=-9\), hence \(y=9\). If \(y=8\), the left side becomes \(2(5)-8=2\), not 1. Exam tip: when \(-y=-9\), change the signs on both sides to get \(y=9\).

Substitute y=3 into the given equation: x+3(3)=15. Thus, x+9=15, so x=6. Choosing 9 is a common error because 9 is the value of 3y, not of x. Exam tip: Always multiply the substituted value by its coefficient carefully.

Adding gives (8x=28), so (x=\frac{7}{2}), which is not among the options. Check calculations carefully.

Substitute \(x=4\) into \(3x+2y=22\): \(3(4)+2y=22\). Thus, \(12+2y=22\), so \(2y=10\) and \(y=5\). If \(y=4\), the left-hand side becomes \(20\), not \(22\). Exam tip: always verify the obtained value in the original equation.

Substituting (y=x-2) gives (3x-2=13), so (x=5) and (y=3). Using the isolated variable makes solving easier.

Subtracting the second equation from the first gives (2x=10), so (x=5). Remove equal (y) terms by subtraction.

From the second equation (x=y+4); substituting gives (4y+4=16), so (y=3) and (x=7). Choose the simpler equation for substitution.

Subtracting the second equation from the first gives (4x=16), so (x=4) and (y=6). After finding one variable, put it in the smaller equation.

Substitute \(x=5\) in \(2x-3y=1\): \(2(5)-3y=1\), so \(10-3y=1\). Hence, \(-3y=-9\) and \(y=3\). If \(y=4\), the left-hand side becomes \(-2\), so it is not correct. Exam tip: after substitution, handle negative signs carefully while transposing terms.

Substituting (x=2y+1) gives (3y+1=13), so (y=4) and (x=9). Combine like terms after substitution.

Subtracting the second equation from the first gives (3y=12), so (y=4). Subtraction is correct to remove the equal (3x).

Given \(y=6\), substitute it into \(x+2y=17\): \(x+2(6)=17\), so \(x+12=17\). Hence, \(x=17-12=5\). Therefore, \(x=5\) is correct. If \(x=6\), the left-hand side becomes \(6+12=18\), not 17. Exam tip: when one variable is given, substitute its value into the other equation to find the remaining variable.

Multiplying (3x+y=14) by (2) gives (6x+2y=28). Recognize equivalent equations formed by multiplication.

Subtracting the equations gives (x-y=0), so (x=y); substituting gives (3x=12). When equality is found, put it into the original equation.

Putting (y=x+4) gives (2x+4=18), so (x=7) and (y=11). In exams, put the expression directly into the other equation.

From the second equation, \(x-y=2\), we get \(y=x-2\). Substituting this into \(2x+y=16\) gives \(2x+(x-2)=16\), so \(3x=18\) and hence \(x=6\). Then \(y=x-2=4\). Therefore, the solution is \((6,4)\). For \((5,6)\), \(x-y=-1\), so it does not satisfy the second equation. Exam tip: verify the ordered pair in both original equations.

Adding both equations gives (2x=16), so (x=8) and (y=2). Add opposite (2y) terms to eliminate them.

Given \(y=10-x\) and \(x=4\), substitute 4 for \(x\): \(y=10-4=6\). Hence, the correct answer is 6. The value 7 would result only if \(x=3\). Exam tip: after substitution, check the subtraction sign carefully.

Substituting x=3 in 3x-y=7 gives 3(3)-y=7, or 9-y=7. Subtracting 9 from both sides gives -y=-2, so y=2. If y=3, then 9-3=6, not 7. In exams, take care of the sign when solving for a term with a negative coefficient.