Algebraic methods: Substitution method and Elimination method.

Multiplying (2x+y=13) by (3) gives (6x+3y=39). Recognizing equivalent equations is also useful in exams.

The governing concept is substitution for a pair of linear equations. From x + y = 12, isolate x to obtain x = 12 − y. Substitute this expression into x − 2y = 3: (12 − y) − 2y = 3. Combining like terms gives 12 − 3y = 3. Subtracting 12 from both sides gives −3y = −9, and dividing by −3 gives y = 3. Hence option B is correct. For verification, x = 12 − 3 = 9. The original equations then become 9 − 2(3) = 3 and 9 + 3 = 12, so both are satisfied exactly. If y were 2, 4, or 5, the resulting x from x+y=12 would not satisfy the first equation.

Subtracting the second equation from the first gives (3x=15), so (x=5) and (y=5). Put the found variable into the smaller equation.

The direct answer is option C: y=13. The first equation already gives y=3x-2. Put this expression into x+y=18: x+(3x-2)=18, so 4x-2=18, 4x=20, and x=5. Now calculate y=3(5)-2=15-2=13. Check: 5+13=18 and 13=3(5)-2. Option C is correct because it satisfies both equations. Option A, y=10, would require x=4 from x+y=18, but then 3x-2=10, not 10? Actually 3(4)-2=10, so it satisfies the first relation but the pair gives x=4 and y=10, and x+y=14, not 18. Option B, y=11, gives x=7 from the sum, but 3(7)-2=19. Option D, y=15, gives x=3 from the sum, but 3(3)-2=7. Thus only C works. Exam cue: substitute the given expression carefully and solve the resulting one-variable equation.

The direct answer is option D: x=7 and y=3. First simplify 5x+5y=50 by dividing every term by 5; this gives x+y=10. The second equation is x-y=4. Add the two equations: (x+y)+(x-y)=10+4, so 2x=14 and x=7. Put x=7 into x+y=10: y=3. Option D therefore satisfies both equations. Option A gives x+y=10 but x-y=2, not 4. Option B gives x+y=10 but x-y=0. Option C gives x+y=10 but x-y=6. Thus those choices fail the second equation, while D passes both. Reducing common factors before elimination makes the arithmetic easier and lowers the chance of error. Remember that dividing an entire equation by the same nonzero number does not change its solution.

The governing concept is substitution in a pair of linear equations. From x − y = 2, solve for x: x = y + 2. Substitute this into 2x + 4y = 28: 2(y + 2) + 4y = 28. Expanding and combining like terms gives 2y + 4 + 4y = 28, so 6y = 24 and y = 4. Substituting back into x = y + 2 gives x = 6. Therefore option A is correct. Verification shows 2(6) + 4(4) = 12 + 16 = 28 and 6 − 4 = 2. The pair in option B reverses the values and fails the equations, while options C and D also fail at least one original relation. Thus only option A is valid.

Given \(x=8\), substitute it in \(x+2y=18\): \(8+2y=18\). Thus, \(2y=10\), so \(y=5\). If \(y=4\), the left-hand side becomes \(16\), not \(18\). Exam tip: after substitution, transpose the constant term first and then divide by the coefficient of the variable.

Given \(3x-y=18\) and \(x=7\), substitute the value of \(x\): \(3(7)-y=18\), so \(21-y=18\). Hence, \(-y=-3\), giving \(y=3\). If \(y=4\), the left-hand side becomes \(21-4=17\), not 18. Exam tip: After substitution, take special care with signs when moving a negative term across the equation.

Subtract \(2x+y=18\) from \(2x+3y=26\): \((2x+3y)-(2x+y)=26-18\Rightarrow 2y=8\Rightarrow y=4\).

Given \(x-y=8\) and \(y=5\), substitute \(y=5\) into the first equation: \(x-5=8\). Adding 5 to both sides gives \(x=13\). For example, \(x=12\) gives \(12-5=7\), not 8. Exam tip: after substitution, carefully use inverse operations to isolate the unknown.

The given equation is \(x+y=20\). Substituting \(y=8\) gives \(x+8=20\). Subtracting 8 from both sides gives \(x=12\). If \(x=10\), the sum would be 18, so it is incorrect. Exam tip: Substitute the given value into the equation and simplify to find the unknown.

The governing concept is elimination. The coefficients of y are −1 and +1, so adding the two equations cancels y immediately: (5x − y) + (x + y) = 25 + 11. This simplifies to 6x = 36. Dividing both sides by 6 gives x = 6, so option C is correct. To verify, substitute x = 6 into x + y = 11; this gives y = 5. The first equation then becomes 5(6) − 5 = 30 − 5 = 25, which is true. Thus the pair is (6,5). Values 4, 5, and 7 would lead to inconsistent y-values or fail the sum obtained after elimination, so they cannot be solutions of the given pair.

The first relation gives x directly in terms of y, so substitute it into the second equation. This replaces x and leaves one equation with y only. Care is needed when combining the two y terms: the expression contains \(-3y\) from the substitution and another \(+y\) from the second equation, so together they make \(-2y\).

Substitute \(x=14-3y\) into \(x+y=8\): \((14-3y)+y=8\). Simplifying gives \(14-2y=8\). Subtract 14 from both sides: \(-2y=-6\). Dividing by -2 gives \(y=3\). A check gives x=14-9=5 and \(5+3=8\). Therefore option D is correct. The negative sign must be retained throughout the calculation.

The two equations describe two conditions on the same values of x and y. Since the coefficient of y is \\(-1\\) in both equations, subtraction is useful: the y terms cancel. This leaves an equation containing only x, which can be solved directly. This is the elimination method for a pair of linear equations.

Subtract the second equation from the first: \\(4x-y\\)-\\(x-y\\)=22-4\\). Simplifying gives \\(4x-y-x+y=18\\), so \\(3x=18\\). Dividing by 3 gives \\(x=6\\). Substitution also confirms the result: the second equation becomes \\(6-y=4\\), so \\(y=2\\), and the first gives \\(24-2=22\\). Therefore option A is correct.

Substitute \(y=2x\) into \(2x+y=16\): \(2x+2x=16\), so \(4x=16\). Hence \(x=4\) and \(y=2\times4=8\). For example, \(x=3,\ y=6\) gives \(2x+y=12\), not 16. Exam tip: verify the obtained values in both original equations.

From the second equation, \(y=6\). Substituting it into \(x+3y=23\) gives \(x+3(6)=23\), or \(x+18=23\). Hence, \(x=5\). The close distractor \(x=6\) is incorrect because it makes the left-hand side \(24\), not \(23\). Exam tip: Verify the obtained value by substituting it back into the original equation.

The substitution method uses one equation that already expresses one variable in terms of the other. Replacing that variable in the second equation changes the pair into one equation with only one unknown. Solve it first, then use the result in the simple expression to find the other variable. Finally, checking both original equations is a useful way to confirm that the ordered pair is correct.

The first equation gives 
m y=x+2
m. Substitute this in 
m 3x+y=18
m: 
m 3x+(x+2)=18
m. Combining like terms gives 
m 4x+2=18
m, so 
m 4x=16
m and 
m x=4
m. Then 
m y=x+2=4+2=6
m. Therefore option D, 
m (x,y)=(4,6)
m, is correct. It satisfies both equations.

The governing concept is elimination of a common variable. Since y has the same coefficient, +1, in both equations, subtract the second equation from the first: (5x + y) − (3x + y) = 31 − 19. The y terms cancel, leaving 2x = 12. Dividing by 2 gives x = 6, so option A is correct. Verification confirms the result: substituting x = 6 into 3x + y = 19 gives 18 + y = 19 and hence y = 1. The first equation then gives 5(6) + 1 = 30 + 1 = 31. Therefore both equations are satisfied. The other proposed values 5, 4, and 3 do not make the difference of the left sides equal to 12 and cannot be the common solution.

From the second equation (x=y+4); substituting gives (5y+4=19), so (y=3) and (x=7). Choose the simpler equation for substitution.

Subtracting the second equation from the first gives (5x=30), so (x=6) and (y=5). After finding one variable, find the other immediately.

Here x is already known to be 4, so put this value into the other equation. This changes the two-variable equation into a one-variable equation in y. The coefficient of y is negative, so the sign must be handled carefully when isolating y. Substitution is sufficient; no elimination between two unknown equations is needed.

Replace x by 4 in \(3x-2y=4\): \(3(4)-2y=4\), so \(12-2y=4\). Subtract 12 from both sides to obtain \(-2y=-8\). Divide by -2, giving \(y=4\). Check: \(3(4)-2(4)=12-8=4\), exactly as required. Hence the correct choice is option D. A common error is to divide by 2 without keeping the negative sign, which would produce an incorrect value.

Substituting (x=3y+2) gives (4y+2=18), so (y=4) and (x=14). Put the given relation directly into the main equation.

Both equations contain the identical term \(2x\). Subtracting the second equation from the first cancels this term, leaving an equation only in y. This is an efficient use of the elimination method because the coefficients of x already match exactly; no multiplication of either equation is required.

Compute \((2x+5y)-(2x+y)=31-15\). The x terms cancel, and \(5y-y=4y\), so \(4y=16\). Dividing by 4 gives \(y=4\). For a check, the second equation becomes \(2x+4=15\), so x=\(11/2\); inserting these values into the first gives \(11+20=31\). Thus y=4 and option B is correct. The useful observation is to remove the equal \(2x\) terms directly.

Given y=8, substitute it into x+2y=21: x+2(8)=21, so x+16=21. Subtracting 16 from both sides gives x=5. Option 4 may seem close, but substituting it gives x+2y=20, not 21. Exam tip: after substitution, multiply first and then isolate the unknown term.

The first equation becomes (2x+y=19), and substituting (y=x-5) gives (3x-5=19). Simplifying the equation first saves time.