Algebraic methods: Substitution method and Elimination method.

Direct answer: option C, \(x=3\). Simplify the first equation by dividing every term by 2: \(4x+2y=20\) becomes \(2x+y=10\). The second equation is \(x+y=7\). Subtract the second from the simplified first: \((2x+y)-(x+y)=10-7\), so \(x=3\). Then \(y=4\), confirming both equations. Option C is correct. Option A would give \(x=1\), which fails the equations. Option B gives 2 and also fails. Option D gives 4; although it would make \(y=3\) from the second equation, it does not satisfy the first. A quick method is to remove common factors before elimination.

From the second equation (x=10-y); substituting gives (10-y-2y=1), so (y=3). Watch brackets and signs in substitution.

The second given equation directly states (x+y=9), so the value of (x+y) is 9. There is no need to solve the first equation. Options such as 8 or 10 do not match the stated sum. Exam tip: If the required expression is already given directly in an equation, avoid unnecessary calculations.

Subtracting the second equation from the first gives (2x=10), so (x=5) and (y=4). Subtraction is correct to remove equal (y) terms.

This is a simultaneous linear-equation problem solved by substitution. Since the first equation gives y directly as 2x − 1, replace y in the second equation x + y = 8 with that expression. We get x + (2x − 1) = 8. Combining like terms gives 3x − 1 = 8; adding 1 gives 3x = 9, and dividing by 3 gives x = 3. Substitution into the first equation then gives y = 2(3) − 1 = 6 − 1 = 5. Therefore option C is correct. Verification shows x + y = 3 + 5 = 8 and y = 2(3) − 1 = 5. The values 3, 4, and 6 do not satisfy both original equations with the same x, so they are not valid solutions.

The first equation has a common factor 5 in every term. Dividing the whole equation by 5 does not change its solutions, but it makes the calculation much easier. It becomes \(x+y=5\). The second equation is already simple. Adding these equations is useful because y and -y cancel, leaving only x.

Add \(x+y=5\) and \(x-y=1\): \(2x=6\), so \(x=3\). Put this value into \(x+y=5\): \(3+y=5\), giving \(y=2\). Checking the second equation gives \(3-2=1\), so the ordered pair is \((3,2)\). Thus option C is correct. Reducing the large coefficients first prevents unnecessary arithmetic and makes the elimination clear.

The first equation is (2(x+2y)=18), so it is (2) times the second. Recognizing proportional equations is also important.

Given \(x=4\), substitute it into \(x+2y=14\): \(4+2y=14\). Thus, \(2y=10\), so \(y=5\). If \(y=4\), the left-hand side becomes \(4+2(4)=12\), not 14. Exam tip: When the value of one variable is given directly, substitute it into the other equation to find the remaining variable.

Given \(x=5\), substitute it into \(2x-y=8\): \(2(5)-y=8\), so \(10-y=8\). Subtracting 10 from both sides gives \(-y=-2\), hence \(y=2\). For example, if \(y=3\), the left side becomes \(10-3=7\), not 8. Exam tip: verify the obtained value by substituting it back into the original equation.

The equations are \\(2x+3y=17\\) and \\(2x+y=9\\). Their x-terms are identical, so subtract the second equation from the first: \\((2x+3y)-(2x+y)=17-9\\). The \\(2x\\) terms cancel, leaving \\(2y=8\\). Dividing by 2 gives \\(y=4\\). Therefore option C is correct. This is the elimination method because one variable is removed by subtraction.

Substitution provides a check. From the second equation, \\(2x=9-y\\). With \\(y=4\\), we get \\(2x=5\\), and the first equation becomes \\(5+12=17\\), which is true. The other listed values for y do not satisfy both equations simultaneously. Hence the supplied answer C is consistent and the cancellation step gives the value efficiently.

Given \(y=2\), substitute it in \(x-y=5\): \(x-2=5\). Adding 2 to both sides gives \(x=7\), so option C is correct. For example, if \(x=6\), then \(6-2=4\), not 5. Exam tip: verify the obtained value by substituting it back into the original equation.

The second equation gives \(y=6\). Substituting it into \(x+y=15\) gives \(x+6=15\). Subtracting 6 from both sides gives \(x=9\), so option C is correct. If \(x=8\), the sum would be 14, not 15. Exam tip: verify the obtained value in the original equation.

The governing concept is the elimination method for a pair of linear equations. The coefficients of y are −1 and +1, so adding the equations cancels y immediately: (3x − y) + (x + y) = 12 + 8. This gives 4x = 20, and division by 4 yields x = 5. Therefore option C is correct. To verify the complete solution, substitute x = 5 into 3x − y = 12: 15 − y = 12, so y = 3. The second equation then becomes 5 + 3 = 8, which is true. If x were 3, 4, or 6, no single corresponding y would satisfy both original equations simultaneously. Thus elimination and substitution both confirm C.

From the first equation, \(x=10-2y\). Substituting it into \(x+y=7\) gives \((10-2y)+y=7\), or \(10-y=7\). Hence, \(y=3\). If \(y=2\), then \(x=6\) and \(x+y=8\), so it does not satisfy the second equation. Exam tip: combine like terms carefully after substitution, especially their signs.

The governing concept is elimination by subtracting equations with identical y terms. Since both equations contain −y, subtract the second from the first: (4x − y) − (x − y) = 13 − 1. Distributing the minus sign gives 4x − y − x + y = 12, so the y terms cancel and 3x = 12. Dividing by 3 gives x = 4, hence option C is correct. Verification is important: from x − y = 1 and x = 4, we obtain 4 − y = 1, so y = 3. Substitution in the first equation gives 4(4) − 3 = 16 − 3 = 13. Therefore the pair satisfies both equations. The other proposed x-values cannot produce one common y for both equations.

Substituting (y=2x) gives (2x+2x=6), so (x=\frac{3}{2}); hence none of the listed pairs would be correct.

The direct answer is option A: \((x,y)=(4,3)\). Using substitution, from \(x-y=1\) we get \(x=y+1\). Put this into \(x+y=7\): \((y+1)+y=7\), so \(2y+1=7\), \(2y=6\), and \(y=3\). Then \(x=y+1=4\). Option A works because \(4+3=7\) and \(4-3=1\). Option B, \((3,4)\), gives sum 7 but difference \(-1\). Option C, \((5,2)\), gives sum 7 but difference 3. Option D, \((2,5)\), gives sum 7 but difference \(-3\). The substitution idea is to express one variable in terms of the other, replace it in the second equation, and then back-substitute. Always check both original equations to catch sign errors.

Substitute \(x=2\) into \(2x+y=8\): \(2(2)+y=8\). Thus, \(4+y=8\), so \(y=4\). If \(y=2\), the left-hand side becomes \(6\), not \(8\). Exam tip: verify the obtained value in the original equation.

Given \(x+y=9\) and \(y=4\), substitute 4 for \(y\) to get \(x+4=9\). Subtracting 4 from both sides gives \(x=5\). Therefore, option C is correct. If \(x=4\), the sum would be 8, not 9. Exam tip: after substitution, perform the same operation on both sides to isolate the unknown.

The governing principle of elimination is to combine equations so that one variable disappears. Here the y coefficients are −1 and +1, so adding the equations is appropriate: (x − y) + (x + y) = 3 + 11. The y terms cancel and produce 2x = 14, so x = 7. Substitute x = 7 into the first equation: 7 − y = 3. Subtracting 7 gives −y = −4, hence y = 4. The ordered solution is therefore (x, y) = (7, 4), making option D correct. Checking the second equation gives 7 + 4 = 11, and the first gives 7 − 4 = 3. The other ordered pairs fail one or both equations.

Given \(3x+y=10\) and \(x=2\), substitute 2 for \(x\): \(3(2)+y=10\), so \(6+y=10\). Subtracting 6 from both sides gives \(y=4\). If \(y=3\), the left-hand side becomes 9, so it is not correct. Exam tip: after substitution, simplify the equation and isolate the unknown term.

Substitute \(x=4\) into \(x+2y=12\): \(4+2y=12\). Subtracting 4 from both sides gives \(2y=8\), so \(y=4\). If \(y=2\), the left-hand side becomes 8, so it is not correct. Exam tip: verify the obtained value in the original equation.

On adding, (y) and (-y) cancel, so (4x=12). In elimination, check signs of like terms carefully.

Given \(x=3\). Substituting it into \(x+y=8\) gives \(3+y=8\), so \(y=5\). Therefore, the solution pair is \((3,5)\). In \((5,3)\), the numbers are present but their order is reversed; an ordered pair must be written as \((x,y)\). Exam tip: always check the order of the coordinates before choosing an answer.

The solution is (x,y)=(3,4), so option A is correct. The first equation already gives y=x+1. Substitute this value of y into x+y=7: x+(x+1)=7. Combining like terms gives 2x+1=7, so 2x=6 and x=3. Then y=x+1=3+1=4. Check both equations: 4=3+1 and 3+4=7, so the pair is valid. Option A, (3,4), satisfies both equations. Option B, (4,3), fails y=x+1 because 3 is not 5. Option C, (2,5), fails the first equation because 5 is not 3. Option D, (5,2), fails both equations: 2 is not 6 and 5+2 is 7 only for the second equation. In substitution, replace one variable by its given expression, solve the remaining variable, and then find the other.