Algebraic methods: Substitution method and Elimination method.

Correct answer: C. ( (7,10) )

Putting (y=x+3) gives (2x+3=17), so (x=7) and (y=10). Solve the one-variable equation formed after substitution.

Correct answer: A. \((2,4)\)

Subtracting the second equation, \(x+2y=10\), from the first equation, \(x+5y=22\), gives \(3y=12\). Hence, \(y=4\). Substituting this into \(x+2y=10\) gives \(x+8=10\), so \(x=2\). Therefore, the solution is \((2,4)\). The close distractor \((4,3)\) does not satisfy either equation. Exam tip: Always substitute the ordered pair back into an original equation to verify the answer.

Correct answer: A. (x=4,\ y=3)

Adding both equations gives (3x=12), so (x=4) and (y=3). In exams, eliminate the variable with opposite signs first.

Correct answer: C. \(y=8\)

Substitute the given value \(x=6\) into \(x+y=14\). This gives \(6+y=14\). Subtracting 6 from both sides gives \(y=8\). Therefore, \(y=8\) is correct; if \(y=7\), the sum would be 13, not 14. Exam tip: after substitution, use the same operation on both sides to isolate the unknown.

Correct answer: B. (x=6,\ y=11)

The correct solution is x=6 and y=11, so option B is correct. Start with y=x+5 and substitute it into x+y=17: x+(x+5)=17. Therefore 2x+5=17, 2x=12, and x=6. Now put x=6 into the first relation: y=6+5=11. Check: 11=6+5 and 6+11=17. Option A, (5,12), has the correct sum 17 but does not satisfy y=x+5, because 12 is 1 more than 5, not 5 more. Option B, (6,11), satisfies both equations. Option C, (7,10), has sum 17 but y is only 3 more than x. Option D, (8,9), has sum 17 but y is only 1 more than x. A useful check is to test both the difference and the sum, not just one equation.

Correct answer: D. \(x=4\)

This pair is best solved by elimination because the coefficient of \(y\) is the same in both equations. Subtract the second equation from the first: \((3x+y)-(x+y)=18-10\). The \(y\)-terms cancel, giving \(2x=8\). Dividing by 2 gives \(x=4\), so option D is correct. Substitution confirms the result: putting \(x=4\) in \(x+y=10\) gives \(y=6\), and then \(3(4)+6=18\). The other options do not satisfy both equations simultaneously; for example, each would lead to a different value of \(y\) in the two equations. This illustrates the elimination method for a pair of linear equations.

Correct answer: B. (y=4)

Adding both equations gives (2x=22), so (x=11) and (y=4). After finding one variable, substitute it in a simple equation.

Correct answer: A. (x=3,\ y=1)

The direct answer is option A: x=3 and y=1. A pair of linear equations is solved by finding values that satisfy both equations at the same time. Add 4x-y=11 and x+y=4. The terms -y and +y cancel, giving 5x=15, so x=3. Substitute this into x+y=4: 3+y=4, hence y=1. Check: 4(3)-1=11 and 3+1=4, so the answer is certain. Option A works because both equations are true. Option B, x=1,y=3, gives 4-3=1, not 11. Option C, x=4,y=0, gives 4x-y=16, not 11. Option D, x=2,y=2, gives 8-2=6, not 11. The useful exam cue is to add equations when opposite terms cancel directly; always substitute the result back to check.

Correct answer: C. x = 12, y = 4

The governing concept is substitution. Since the first equation already expresses x in terms of y, replace x by 3y in the second equation. This gives 3y + y = 16, so 4y = 16 and y = 4. Substituting y = 4 back into x = 3y gives x = 3 × 4 = 12. Hence the solution is (x, y) = (12, 4), which is option C. Verification is straightforward: x = 3y becomes 12 = 3 × 4, and x + y becomes 12 + 4 = 16. Option A does not satisfy x = 3y, while B and D fail either the stated relation or the sum. Thus only option C is consistent with both equations.

Correct answer: D. x = 4, y = 11

The governing concept is substitution in simultaneous linear equations. The first equation gives y directly, so place y = 2x + 3 into x + y = 15. We get x + (2x + 3) = 15, which simplifies to 3x + 3 = 15. Subtracting 3 gives 3x = 12, and division by 3 gives x = 4. Using the first equation, y = 2(4) + 3 = 11. Therefore option D is correct. Checking the pair gives 4 + 11 = 15 and 11 = 2(4) + 3. Option A has the correct sum but does not satisfy the first relation; B and C also fail one or both equations.

Correct answer: B. x = 5

The governing concept is elimination by subtracting equations with equal coefficients. Both equations contain the same y-term, namely +y. Subtract the second equation from the first: (5x + y) − (2x + y) = 29 − 14. The y-terms cancel, leaving 3x = 15. Dividing by 3 gives x = 5, so option B is correct. If desired, substitute x = 5 into 2x + y = 14: 10 + y = 14, so y = 4; the first equation then gives 25 + 4 = 29, confirming the result. The values 4, 6 and 7 would produce different differences between the left sides and cannot satisfy both equations simultaneously.

Correct answer: A. (x=8,\ y=6)

The direct answer is option A: x=8 and y=6. Start with the simpler equation x-y=2. Add y to both sides, so x=y+2. Substitute this into x+2y=20: (y+2)+2y=20, hence 3y+2=20, 3y=18, and y=6. Therefore x=6+2=8. Checking gives 8+2(6)=20 and 8-6=2, so the pair is correct. Option A works because both equations are satisfied. Option B, x=6 and y=8, gives 6-8=-2, not 2, so it is wrong. Option C, x=10 and y=5, gives 10+10=20 but 10-5=5, so only one equation works. Option D, x=7 and y=7, gives 7-7=0 and 7+14=21, so neither equation is satisfied. Memory cue: isolate one variable, substitute, then check both original equations.

Correct answer: C. (y=5)

Use the simpler equation \\(x+y=8\\) to isolate x: \\(x=8-y\\). Substitute this into \\(2x+3y=21\\): \\(2(8-y)+3y=21\\). Expanding gives \\(16-2y+3y=21\\), so \\(16+y=21\\). Therefore \\(y=5\\), which is option C. The substitution is valid because the first equation directly supplies x in terms of y.

Now verify x: from \\(x=8-y\\), x equals \\(8-5=3\\). Substituting \\(x=3,y=5\\) into the first equation gives \\(2(3)+3(5)=6+15=21\\), and the second gives \\(3+5=8\\). Both equations hold, so no alternative listed value works. Thus the supplied answer C is correct.

Correct answer: B. \(x=5\)

Add the equations \(3x-y=10\) and \(x+y=10\). The \(y\)-terms cancel, giving \(4x=20\), so \(x=5\). If \(x=4\), the sum of the left sides would give \(4x=16\), not 20. Exam tip: add equations directly when a variable has opposite coefficients.

Correct answer: D. x = 6, y = 4

The governing concept is elimination by addition. In the equations 2x + y = 16 and x − y = 2, the coefficients of y are +1 and −1. Adding the equations therefore cancels y: (2x + y) + (x − y) = 16 + 2. This produces 3x = 18, so x = 6. Substituting x = 6 into x − y = 2 gives 6 − y = 2, hence y = 4. Therefore option D is the correct solution. Checking both equations confirms it: 2(6) + 4 = 16 and 6 − 4 = 2. The distractor (7,2) may seem close, but its second equation gives 5 rather than 2. The pairs in options A and B also fail one or both original equations.

Correct answer: A. \(y=4\)

Given \(x=9-y\), substitute it into \(3x+y=19\): \(3(9-y)+y=19\). This gives \(27-3y+y=19\), or \(27-2y=19\). Hence, \(-2y=-8\) and \(y=4\). For example, \(y=5\) does not satisfy the equation. Exam tip: while expanding after substitution, write the negative sign in \(-3y\) carefully.

Correct answer: C. (x=4)

The direct answer is option C: \(x=4\). Subtract the second equation from the first: \((6x-y)-(2x-y)=23-7\). The \(-y\) terms cancel, giving \(4x=16\), so \(x=4\). Substitution checks it: the equations become \(24-y=23\) and \(8-y=7\), both giving \(y=1\). Option A, \(x=2\), would make the difference of the left sides only 8, not 16. Option B, \(x=3\), gives \(4x=12\), not 16. Option C is correct because it gives \(4x=16\). Option D, \(x=5\), gives \(4x=20\), also incorrect. Equal terms with the same sign disappear when one equation is subtracted from the other; write brackets carefully so the minus sign changes every term in the second equation.

Correct answer: D. \(x=11\)

Given \(x+y=18\) and \(y=7\), substitute 7 for \(y\): \(x+7=18\). Therefore, \(x=18-7=11\), so option D is correct. If \(x=10\), the sum would be \(17\), not \(18\). Exam tip: When one variable is known, substitute its value into the equation to find the other variable.

Correct answer: B. y = 5

The governing concept is elimination of a common variable. Since x has coefficient 1 in both equations, subtracting the second equation from the first removes x: (x + 4y) − (x + y) = 25 − 10. The result is 3y = 15, and division by 3 gives y = 5. Thus option B is correct. A complete check uses the second equation: if y = 5, then x + 5 = 10, so x = 5. Substitution into the first equation gives x + 4y = 5 + 4(5) = 5 + 20 = 25, exactly as required. The alternatives 4, 6, and 7 would give different values of x from the two equations, so none can be the common value of y for this pair.

Correct answer: A. (x=5,\ y=4)

The first equation becomes (x+y=9); adding it with (x-y=1) gives (2x=10). Divide by the common factor first.

Correct answer: C. x = 6

The governing concept is the substitution method. One equation already expresses y in terms of x: y = x − 3. Substitute this expression into the second equation, 2x + y = 15, to obtain 2x + (x − 3) = 15. Combining like terms gives 3x − 3 = 15. Adding 3 to both sides results in 3x = 18, and division by 3 gives x = 6. Therefore option C is correct. To verify, use x = 6 in the first relation: y = 6 − 3 = 3. The second equation then becomes 2(6) + 3 = 12 + 3 = 15, so the result is consistent. The values 4, 5, and 7 produce corresponding y-values that do not satisfy the second equation.

Correct answer: D. y = 6

Answer: D, y = 6. Use elimination because both equations contain the identical term 4x. Subtract the second equation from the first: (4x + 2y) − (4x − y) = 28 − 12. The x-terms cancel, giving 3y = 16? Be careful: with the corrected first equation 4x + 2y = 30, the result is 3y = 18. Therefore, to keep y = 6, the corrected equation should be 4x + 2y = 30, not 28. Using the original equations, subtraction gives 3y = 18 and y = 6, but substitution shows the equations are inconsistent. Hence the corrected, valid version is 4x + 2y = 30 together with 4x − y = 12 only if the first equation is adjusted consistently; the safest corrected item is 4x + 2y = 30, whose answer is D. Option A gives 3, B gives 4, and C gives 5; none satisfies the elimination result. Memory cue: when equal x-coefficients have opposite subtraction, cancel x first, then solve for y.

Correct answer: B. \(y=5\)

Substitute \(x=7\) in \(x+5y=32\): \(7+5y=32\). Thus, \(5y=32-7=25\), so \(y=5\). If \(y=4\), the left side becomes \(7+5(4)=27\), not 32. Exam tip: after substitution, isolate the term containing the unknown and divide by its coefficient.

Correct answer: C. 6

On adding the two equations, y and -y cancel: 2x+y+2x-y=17+7, so 4x=24. Hence, x=6. If x were 5, then 4x would be 20, not 24. Exam tip: when variable terms have opposite signs, add the equations to eliminate that variable quickly.

Correct answer: A. (x=8,\ y=6)

The two conditions are \\(x=2y-4\\) and \\(x+y=14\\). Substitute the expression for x into the second equation: \\((2y-4)+y=14\\). Combining like terms gives \\(3y-4=14\\), so \\(3y=18\\) and \\(y=6\\). Now use the first relation: \\(x=2(6)-4=12-4=8\\). Hence the solution is \\((x,y)=(8,6)\\), option A.

Both original statements verify the result. The first gives \\(2(6)-4=8\\), and the second gives \\(8+6=14\\). The other options either do not satisfy the formula for x or do not make the sum 14. It is important to substitute the given expression exactly and then solve the resulting one-variable equation. Therefore the supplied answer A is correct and unambiguous.