Algebraic methods: Substitution method and Elimination method.

Multiplying the second equation by (3) gives (15x-9y=42). Add and solve carefully because fractional answers are possible.

Multiply the first equation by (2) to eliminate (y). After finding (x), substitute back before evaluating (x+y).

Adding gives (21x=70), so (x=\frac{10}{3}). Then (y=\frac{14}{3}), hence (y-x=\frac{4}{3}).

Adding gives (16x=48), so (x=3). From the second equation (y=\frac{27}{7}), hence (2x-y=\frac{15}{7}).

Let chair be (c) and table be (t), so (3c+2t=4900), (2c+3t=5600). Elimination gives (t=1400).

Multiply the first equation by (2) to get (4x+10y=62). Adding gives (7x=50), so check fractional values too.

From the first equation, x=18-2y. Substituting this in the second equation gives 4(18-2y)-y=9, or 72-9y=9. Thus, 9y=63 and y=7. Therefore, option B is correct. Exam tip: In the substitution method, express one variable in terms of the other and substitute it into the remaining equation carefully.

Subtracting the second equation from the first gives (12y=60), so (y=5). Then (x=\frac{34}{5}), hence (x-y=\frac{9}{5}).

Let the angles be (x) and (y), so (x+y=90^\circ) and (x-y=28^\circ). Adding gives (2x=118^\circ), so the larger angle is (59^\circ).

Let (u=x-1) and (v=y+1). Solve (3u+2v=48), (2u-3v=-6) and substitute back carefully.

Adding the two equations eliminates the y-terms because their coefficients are opposites: (7x+11y)+(14x-11y)=103+23, giving 21x=126. Hence, x=126/21=6, so option C is correct. Exam tip: look first for equal and opposite coefficients, as they allow immediate elimination of one variable.

Adding gives (9x=60), so (x=\frac{20}{3}). Substitute back carefully to avoid arithmetic errors.

Substituting (y=2x+3) gives (5x-2(2x+3)=1). This gives (x=7); handle the negative sign outside brackets carefully.

Adding gives (24x=72), so (x=3). From the second equation (y=\frac{23}{7}), so (x+2y=\frac{67}{7}).