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.

The governing concept is substitution because y is already isolated in the first equation. Replace y in the second equation by x − 2: 2x + (x − 2) = 13. Combining like terms gives 3x − 2 = 13. Add 2 to both sides to obtain 3x = 15, then divide by 3 to get x = 5. Substitute this value into y = x − 2: y = 5 − 2 = 3. Therefore the solution is (x, y) = (5, 3), so option A is correct. Verification gives 2(5) + 3 = 13 and 3 = 5 − 2. Option C has the values in an incompatible combination, while B and D fail at least one original equation.

The two equations are \\(4x+y=21\\) and \\(2x+y=11\\). Both contain the same term \\(y\\) with the same coefficient. Subtracting the second equation from the first removes \\(y\\): \\(4x+y-(2x+y)=21-11\\). This gives \\(2x=10\\), and dividing both sides by 2 gives \\(x=5\\). Therefore, option D is correct.

A quick check confirms the result. Substituting \\(x=5\\) into the second equation gives \\(2(5)+y=11\\), so \\(y=1\\). The first equation then gives \\(4(5)+1=21\\), which is also true. The key method is elimination: equal terms with equal coefficients cancel when one equation is subtracted from the other.

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.

One equation already gives x in terms of y: \(x=2y+1\). This makes substitution the natural method. Replace x by \(2y+1\) in the second equation. Then there is only one unknown, y, so it can be found with ordinary algebra. After finding y, substitute it back into the first relation to find x.

Using substitution, \(x+y=13\) becomes \((2y+1)+y=13\). Combine like terms to get \(3y+1=13\), then subtract 1: \(3y=12\). Dividing by 3 gives \(y=4\). Now \(x=2(4)+1=9\). Checking gives \(9+4=13\), so the solution is \((9,4)\), which is option C. The other choices do not satisfy both equations.

The governing concept is the elimination method for a pair of linear equations in two variables. Both equations contain the term 3x, so subtracting the second equation from the first removes x without changing the common solution. Thus, (3x + 4y) − (3x + y) = 25 − 13, which simplifies to 3y = 12. Dividing both sides by 3 gives y = 4. To verify the result, substitute y = 4 into the second equation: 3x + 4 = 13, so x = 3. The first equation then becomes 3(3) + 4(4) = 9 + 16 = 25, confirming the solution. Therefore, option C is correct. Values 2, 3, and 5 do not produce a common x-value satisfying both original equations.

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.

The governing idea is equivalence of equations and the effect of multiplying every term by the same non-zero number. Starting with 3x + y = 14, multiply the left side and the right side by 2: 2(3x + y) = 2(14). This becomes 6x + 2y = 28, which is exactly the first equation. Therefore the first equation is twice the second, so both equations represent the same line and do not produce two different independent equations. Option A states this accurately. Option B is false because equivalent equations do not necessarily give different solutions. Options C and D are false because x is present and both equations have constants.

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.

The governing concept is elimination in simultaneous linear equations. The coefficients of y are +2 and −2, so adding the two equations eliminates y directly: (x + 2y) + (x − 2y) = 12 + 4. This gives 2x = 16 and therefore x = 8. Substituting x = 8 into x + 2y = 12 gives 8 + 2y = 12, so 2y = 4 and y = 2. Hence the ordered pair is (8, 2), which is option C. Verification is important: 8 + 2(2) = 12 and 8 − 2(2) = 4, so both original equations hold. The other pairs fail at least one equation; for example, (6,3) gives 12 in the first equation but 0 in the second, not 4.

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.

Putting (x=2y) gives (3y=15), so (y=5) and (x=10). Substitution is easiest when a ratio relation is given.

Subtracting the second equation from the first gives \((2x+5y)-(2x+y)=29-13\). Thus, \(4y=16\), so \(y=4\). The close distractor \(y=3\) would require \(4y=12\), which is not obtained here. Exam tip: while subtracting equations, carefully keep track of the signs of all terms.

Given \(y=2\), substitute it into \(x-3y=1\): \(x-3(2)=1\), so \(x-6=1\). Adding 6 to both sides gives \(x=7\). If \(x=8\), the left-hand side becomes \(8-6=2\), not 1. Exam tip: after substitution, simplify brackets and negative signs carefully.

Adding the two equations gives \((3x+2y)+(3x-2y)=20+4\). The \(+2y\) and \(-2y\) terms cancel, so \(6x=24\). Hence, \(x=4\). \(x=6\) may seem close, but it results from solving \(6x=24\) incorrectly. Exam tip: in elimination, first check whether the coefficients of one variable are opposites.

Given y=12-2x and x=5, substitute 5 for x: y=12-2(5)=12-10=2. Hence, the correct answer is 2. A value such as 3 can result from an error in multiplication or subtraction. Exam tip: after substitution, perform multiplication before addition or subtraction.