Algebraic methods: Substitution method and Elimination method.

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

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.

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.