Algebraic methods: Substitution method and Elimination method.

Putting (x=5-y) gives (4(5-y)-5y=2), so (y=2) and (x=3). Be careful while combining negative terms.

Subtracting the second equation from the first gives (5y=25), so (y=5) and (x=8). Eliminating equal (5x) terms is the right step.

From the second equation, y = 4x - 3. Substituting this into the first equation gives 2x + 3(4x - 3) = 33, so 14x = 42 and x = 3. Then y = 4(3) - 3 = 9. Hence, the solution is (3, 9). Option (5, 6) satisfies the second equation, but it gives 2x + 3y = 28, not 33. Exam tip: Always verify the ordered pair in both original equations.

From the second equation, \(3x-y=15\), so \(y=3x-15\). Substituting this into \(x+5y=37\) gives \(x+5(3x-15)=37\). Hence, \(16x=112\), so \(x=7\) and \(y=6\). Therefore, \((7,6)\) is the correct solution. For example, \((8,5)\) gives \(33\), not \(37\), in the first equation. Exam tip: always verify the ordered pair in both equations.

Adding both equations gives (14x=56), so (x=4) and (y=7). Add opposite (2y) terms to eliminate them.

From the first equation \(2x-y=8\), we get \(y=2x-8\). Substituting this into the second equation gives \(3x+2(2x-8)=47\), so \(7x=63\) and \(x=9\). Then \(y=2(9)-8=10\). Hence, the solution is \((9,10)\). Option \((8,8)\) satisfies the first equation but not the second. Exam tip: always verify the ordered pair in both original equations.

Subtracting the second from the first gives (3x=27), so (x=9) and (y=5). Remove equal (3y) terms to solve quickly.

Multiplying the first equation by (2) and the second by (3) eliminates (y). The solution is (x=3,\ y=4).

From the first equation, \(y=4x-17\). Substituting this into the second equation gives \(2x+3(4x-17)=19\). Thus, \(14x-51=19\), so \(14x=70\) and \(x=5\). The nearby option \(x=4\) does not satisfy both equations together. Exam tip: after substitution, multiply 3 by both \(4x\) and \(-17\) while expanding the bracket.

Multiply the first equation by 2 to get \(10x+4y=48\). Subtract \(3x+4y=22\) from it: \(7x=26\), so \(x=\frac{26}{7}\). Substituting this into \(5x+2y=24\) gives \(2y=24-\frac{130}{7}=\frac{38}{7}\), hence \(y=\frac{19}{7}\). The distractor \(y=2\) does not satisfy the original equations. Exam tip: make one variable’s coefficients equal before subtracting to eliminate it quickly.

From the second equation, \(y=3x-8\). Substituting this into \(x+2y=11\) gives \(x+2(3x-8)=11\), so \(7x=27\) and \(x=\frac{27}{7}\). Then \(y=3\times\frac{27}{7}-8=\frac{25}{7}\). Hence, \(x+y=\frac{27}{7}+\frac{25}{7}=\frac{52}{7}\). The option \(8\) may result from assuming integer values, but the given equations have fractional solutions. Exam tip: when adding fractions with the same denominator, add only their numerators.

From the second equation, \(y=9-2x\). Substituting this into \(7x-3y=20\) gives \(7x-3(9-2x)=20\), so \(13x=47\). Hence, \(x=\frac{47}{13}\) and \(y=9-2\times\frac{47}{13}=\frac{23}{13}\). Although \((3,3)\) satisfies the second equation, it gives \(7(3)-3(3)=12\neq20\) in the first equation. Exam tip: always verify the ordered pair in both original equations.

Multiply the second equation by (5) and add with the first. Checking the options in both equations gives the correct value (y=5).

The coefficients of y are +5 and -5, so y is eliminated when the equations are added: \(6x+5y+4x-5y=39+11\). Thus, \(10x=50\), giving \(x=5\). If \(x=4\), then \(10x=40\), so it cannot satisfy the added equation. Exam tip: Before eliminating, look for terms with equal coefficients and opposite signs.

Adding the two equations eliminates the y-terms: 8x=40, so x=5. Substituting x=5 into 3x+2y=23 gives 15+2y=23, hence y=4. Therefore, x-y=5-4=1. Option 3 is incorrect because it does not match the difference of the correct values of x and y. Exam tip: Add equations when a variable has equal coefficients with opposite signs.

Multiplying the first equation by -2 gives (-4x-6y=-26). On adding it to (4x-5y=1), we get (-4x+4x=0), so x is eliminated. Multiplying by 2 would produce (4x), which would not cancel with the (4x) in the second equation when added. Exam tip: for elimination by addition, make the coefficients equal in magnitude and opposite in sign.

Use (y=2x-3) from the second equation. Substitution and option checking show (x=4) is correct.

The second equation is (2) times the first. Hence both represent the same line and have infinitely many solutions.

The first equation becomes (x+2y=6), while the second is (x+2y=8). Same left side with different right side gives no solution.

Let the larger number be \(x\) and the smaller number be \(y\). Then \(x+y=21\) and \(x-y=5\). Adding the equations gives \(2x=26\), so \(x=13\). Therefore, the larger number is 13. If 14 were chosen, the other number would be 7 and their difference would be 7, not 5. Exam tip: In sum-and-difference questions, add the two equations to find the larger number directly.