Algebraic methods: Substitution method and Elimination method.

The direct answer is option B: (x,y)=(4,3). In substitution, first isolate a variable from the easier equation. From x-y=1, subtract x or rearrange to get y=x-1. Substitute into 2x+y=11: 2x+(x-1)=11, so 3x-1=11, 3x=12, and x=4. Then y=4-1=3. Checking gives 2(4)+3=11 and 4-3=1. Option B works because both equations hold. Option A, (3,5), gives 2(3)+5=11 but 3-5=-2, so it fails the second equation. Option C, (5,1), gives 10+1=11 but 5-1=4, not 1. Option D, (2,7), gives 4+7=11 but 2-7=-5. Therefore B is the only valid pair. Memory cue: when a coefficient is 1 or -1, isolate that variable first; always verify both original equations.

From the second equation, \(x-y=2\), so \(x=y+2\). Substituting this into \(2x+3y=19\) gives \(2(y+2)+3y=19\), or \(5y=15\). Hence, \(y=3\), and then \(x=3+2=5\). Therefore, the solution is \((5,3)\). For example, \((6,2)\) gives \(x-y=4\), so it is not correct. Exam tip: verify the ordered pair in both original equations.

Subtracting the second equation from the first gives (3y=12), so (y=4) and (x=5). Equal (x)-coefficients make elimination faster.

Adding (x+y=14) and (x-y=4) gives (2x=18), so (x=9) and (y=5). In word problems, form equations first.

From the second equation, 4x-y=5 ⇒ y=4x-5. Substituting this into the first equation gives 2x+3(4x-5)=13. Thus, 14x-15=13, so 14x=28 and x=2. Now, y=4(2)-5=3. Therefore, the solution is (2, 3). Option (3, 2) does not satisfy the second equation because 4(3)-2=10, not 5. Exam tip: Always verify the obtained pair in both original equations.

The governing concept is elimination of one variable from simultaneous linear equations. Both equations have the same y-term, 2y. Subtract the second equation from the first: (5x + 2y) − (3x + 2y) = 34 − 22. The y terms cancel, giving 2x = 12 and therefore x = 6. Substitute this into 3x + 2y = 22: 3(6) + 2y = 22, so 18 + 2y = 22, 2y = 4, and y = 2. Hence the solution is (6,2), option C. Checking gives 5(6)+2(2)=34 and 3(6)+2(2)=22. Every other listed ordered pair fails at least one original equation.

From (x-y=2), (x=y+2), so (\frac{y+2}{2}+y=7) and (y=4). In fraction questions, try to clear denominators first.

Putting (x=10-y) gives (3(10-y)+2y=24), so (y=6) and (x=4). A sum equation makes substitution easier.

The governing concept is the elimination method for a pair of simultaneous linear equations. Because the coefficients of y are both 5, subtract the first equation from the second: (3x + 5y) − (2x + 5y) = 29 − 26. The y terms cancel, leaving x = 3. Substitute x = 3 into 2x + 5y = 26: 2(3) + 5y = 26, so 6 + 5y = 26, 5y = 20, and y = 4. Therefore the ordered solution is (3, 4), so option D is correct. Verification gives 2(3) + 5(4) = 26 and 3(3) + 5(4) = 29. The other ordered pairs fail at least one original equation.

Subtract the first equation from the second: \(3x+2y-(x+2y)=25-11\). Thus, \(2x=14\), so \(x=7\). Substituting \(x=7\) into \(x+2y=11\) gives \(2y=4\), hence \(y=2\). Therefore, the solution is \((7, 2)\). Option B, \((5,3)\), does not satisfy the second equation. Exam tip: when a variable has equal coefficients in both equations, subtract the equations to eliminate it directly.

This is a substitution problem involving two linear equations. The relation x = 3y − 1 can be substituted into x + y = 11. Therefore (3y − 1) + y = 11. Combining like terms gives 4y − 1 = 11, so 4y = 12 and y = 3. Now substitute y = 3 into x = 3y − 1: x = 3(3) − 1 = 8. Hence the ordered pair (x, y) is (8, 3), which is option C. Verification is important because an ordered pair has a fixed order: 8 + 3 = 11 and 8 = 3(3) − 1. The other alternatives either have the wrong sum or violate the given expression for x.

These are simultaneous linear equations, and substitution gives a systematic solution. From x + y = 13, isolate x as x = 13 − y. Substitute this into 2x + 3y = 34: 2(13 − y) + 3y = 34. Expanding and combining like terms gives 26 − 2y + 3y = 34, so y = 8. Now substitute y = 8 into x + y = 13: x + 8 = 13, hence x = 5. Therefore the solution is (x, y) = (5, 8), making option A correct. Verification gives 5 + 8 = 13 and 2(5) + 3(8) = 10 + 24 = 34. Option B reverses the coordinates, and C and D fail the equations.

From the second equation, \(2x-y=9\), we get \(y=2x-9\). Substituting this into the first equation gives \(4x+2(2x-9)=30\), so \(8x=48\) and \(x=6\). Then \(y=2(6)-9=3\). Hence, the solution is \((6,3)\). The close option \((5,5)\) does not satisfy the second equation because \(2(5)-5=5\), not 9. Exam tip: substitute the final values into both original equations to verify the solution.

The governing concept is elimination of opposite terms in a pair of simultaneous linear equations. Here the coefficients of y are −2 and +2, so adding the equations removes y without first solving for either variable: (7x − 2y) + (7x + 2y) = 41 + 57. Thus 14x = 98 and x = 7. Substitute x = 7 into 7x + 2y = 57: 49 + 2y = 57, so 2y = 8 and y = 4. Therefore the ordered pair is (7, 4), making option C correct. Verification gives 49 − 8 = 41 and 49 + 8 = 57 in the original equations. The other ordered pairs fail at least one equation and therefore cannot be the simultaneous solution.

The direct answer is option B: infinitely many solutions. Compare the equations: multiplying 2x+3y=12 by 2 gives 4x+6y=24, exactly the second equation. Therefore the two equations are not two different lines; they are the same line written in different forms. Every point on that line satisfies both equations, so there are infinitely many ordered pairs, not just one pair. Option B is correct. Option A, no solution, would occur when the left-side coefficients are proportional but the constants are not proportional, producing parallel different lines. That does not happen here. Option C, exactly one solution, occurs when the two lines intersect once; here they coincide. Option D, two different solutions, is not the standard result for two linear equations representing the same line; in fact there are infinitely many. Memory cue: if one complete equation is a non-zero multiple of the other, write “infinitely many solutions.”

Multiplying the first equation by (2) gives the same left side but constant (16). Since it conflicts with (20), there is no solution.

Subtracting the first from the second gives (3x=12), so (x=4) and (y=4). After elimination, substitute into any original equation.

From the first equation, \(x=2y+1\). Substituting this into \(3x-y=18\) gives \(3(2y+1)-y=18\). Thus, \(6y+3-y=18\), so \(5y=15\) and \(y=3\). Hence, \(x=2(3)+1=7\), making \((7,3)\) the correct pair. Option \((5,2)\) satisfies the first equation but not the second one. Exam tip: verify the obtained values in both original equations.

Putting (3x-5) in place of (y) gives (5x-5=20), so (x=5) and (y=10). Replace the entire expression carefully.

If the father's age is (x) and the son's age is (y), then (x+y=50) and (x=4y). This gives (5y=50), so the ages are (40) and (10).

The direct answer is option C: \((x,y)=(7,5)\). The first equation gives \(x+y=12\). Divide the second equation by 10: \(5x+3y=50\). From the first equation, \(y=12-x\). Substitute: \(5x+3(12-x)=50\), so \(5x+36-3x=50\), \(2x=14\), and \(x=7\). Then \(y=12-7=5\). Option A, \((6,6)\), has the correct total but gives ticket value 480, not 500. Option B, \((8,4)\), totals 12 but gives 520. Option C gives total 12 and \(50(7)+30(5)=350+150=500\), so it is correct. Option D, \((5,7)\), gives 480. A useful check is to test both the count equation and the money equation, not just one.

On adding the two equations, the y-terms cancel: (x+2y)+(3x-2y)=13+23. Thus, 4x=36 and x=9. Substituting x=9 into x+2y=13 gives 9+2y=13, so y=2. Therefore, the solution is (9,2). Option (8,3) gives 14 in the first equation, so it is not correct. Exam tip: In elimination, add or subtract equations to cancel terms with opposite coefficients.

From the first equation, y = 13 - 2x. Substituting this into the second equation gives x + 3(13 - 2x) = 19, so -5x = -20 and x = 4. Then y = 13 - 2(4) = 5. Hence, the solution is (4, 5). The close distractor (5, 4) gives 2x + y = 14, so it is not correct. Exam tip: Always verify the ordered pair in both original equations.

Putting (x=y+6) gives (\frac{y+6}{3}+y=6), so (y=3) and (x=9). Matching denominators reduces mistakes in fractions.

From the first equation, \(y=2x-7\). Substituting this into \(x+2y=16\) gives \(x+2(2x-7)=16\), so \(5x=30\) and hence \(x=6\). Then \(y=2(6)-7=5\). Therefore, the solution is \((6,5)\). For \((5,6)\), \(2x-y=4\), so it does not satisfy the first equation. Exam tip: always substitute the ordered pair into both original equations to verify the answer.