Simultaneous Equations

Substitute $y = x + 2$ into $x + y = 8$: $x + (x + 2) = 8$, so $2x + 2 = 8$, $2x = 6$ and $x = 3$. Then $y = 3 + 2 = 5$. ($x = 5,\; y = 3$ swapped $x$ and $y$; $x = 4,\; y = 4$ split 8 equally, ignoring $y = x + 2$.)

Scale to match $x$: $(3x+2y=7)\times 2$ and $(2x+5y=12)\times 3$ give $6x+4y=14$ and $6x+15y=36$. Subtract: $11y = 22$, so $y = 2$, then $3x + 4 = 7$ gives $x = 1$.

Let the boat speed be $b$ and the current $c$. Downstream: $b + c = \dfrac{30}{2} = 15$; upstream: $b - c = \dfrac{30}{3} = 10$. Add the equations: $2b = 25$, so $b = 12.5$ km/h (and $c = 2.5$ km/h). ($2.5$ km/h is the current speed.)

Rearrange the first as $y = 10 - x$ and substitute into the second: $2x - (10 - x) = 5$. Distribute the minus: $2x - 10 + x = 5$, so $3x - 10 = 5$, $3x = 15$ and $x = 5$, then $y = 10 - 5 = 5$. ($x = 15,\; y = -5$ did not distribute the minus, solving $2x - 10 - x = 5$.)

At the intersection both $y$-values are equal: $2x - 1 = -x + 5$, so $3x = 6$ and $x = 2$. Then $y = 2(2) - 1 = 3$, giving the point $(2, 3)$. (The intersection of two lines is the solution of their simultaneous equations.)

Eliminate $y$ by adding the equations: $2x = 10$, so $x = 5$. Substitute back: $5 + y = 7$, so $y = 2$. ($x = 10,\; y = -3$ did not divide $2x = 10$; $x = 2,\; y = 5$ subtracted instead of adding and swapped values.)

The $y$-coefficients are equal, so eliminate $y$ by subtracting: $(2x + y) - (x + y) = 10 - 6$, giving $x = 4$. Substitute back: $4 + y = 6$, so $y = 2$. ($x = 2,\; y = 4$ swapped the values.)

Let nuts $= n$ and raisins $= r$ (in kg): $n + r = 10$ and $8n + 5r = 62$ (since $10 \times 6.20 = 62$). Substitute $r = 10 - n$: $8n + 5(10 - n) = 62$, so $3n + 50 = 62$, giving $3n = 12$ and $n = 4$ kg. ($6$ kg is the amount of raisins.)

Multiply the first by 3 and the second by 2 to match the $x$-coefficients: $6x + 9y = 39$ and $6x + 4y = 24$. Subtract: $5y = 15$, so $y = 3$, then $2x + 9 = 13$ gives $x = 2$. ($x = 3,\; y = 2$ swapped the values, a common trap when both equations look similar.)

Let the tens digit be $t$ and the units digit $u$: $t + u = 9$. The number is $10t + u$ and the reversed number is $10u + t$, so $(10u + t) - (10t + u) = 27$ gives $9u - 9t = 27$, that is $u - t = 3$. Solving $t + u = 9$ and $u - t = 3$: $u = 6$, $t = 3$, so the number is $36$. ($63$ is the reversed number.)

Test each pair in both equations. For $(3, 1)$: $2(3) + 1 = 7$ and $3 - 1 = 2$, so both equations hold and this is the solution. ($(1, 5)$ satisfies only the first; $(4, 2)$ satisfies only the second.) The classic trap: each distractor satisfies one equation but not both, so you must check both.

Both equations have the same left-hand side $2x + y$ but different right-hand sides (5 and 8). No pair can make $2x + y$ equal both 5 and 8, so the lines are parallel and never meet: there is no solution. (Infinite solutions would require the two equations to be the same line; two linear equations can never have exactly two solutions.)

The second equation is exactly $2 \times$ the first: $2(x + y) = 2(3)$ gives $2x + 2y = 6$. The two equations are the same line, so every point on it satisfies both: there are infinitely many solutions. (No solution would require parallel non-identical lines; one solution would require two distinct intersecting lines.)

The point of intersection of two lines is the solution to the simultaneous equations: both equations are satisfied when $x = 4$ and $y = 2$. ($x = 2,\; y = 4$ swapped the coordinates; in $(x, y)$ notation the first coordinate is $x$. A solution must specify both variables, so "$x = 4$ only" is incomplete.)

When one equation is already in the form $y = \ldots$, substitution is the most efficient first step. Substitute $y = 2x - 1$ into the other equation: $3x + 4(2x - 1) = 17$, so $3x + 8x - 4 = 17$ and $11x = 21$. (Adding or subtracting the equations as written would not eliminate either variable because the coefficients do not match; graphing is slow and imprecise.)

Let the numbers be $x$ and $y$ with $x > y$: $x + y = 14$ and $x - y = 4$. Add the equations: $2x = 18$, so $x = 9$, then $y = 5$. (7 and 7 split 14 equally, but two equal numbers cannot differ by 4; 10 and 4 differ by 6.)

Let chocolate $= c$ and muesli $= m$: $2c + 3m = 11$ and $4c + m = 12$. Multiply the second by 3: $12c + 3m = 36$. Subtract the first: $10c = 25$, so $c = 2.50$. ($2.00 is the muesli price, not the chocolate.)

Let Sarah $= s$ and brother $= b$: $s + b = 25$ and $s - 5 = 2(b - 5)$, which gives $s = 2b - 5$. Substitute: $(2b - 5) + b = 25$, so $3b = 30$, $b = 10$ and $s = 15$. (10 is the brother's age, not Sarah's.)

Let the smaller angle $= x$ and the larger $= y$. Supplementary: $x + y = 180$. Relationship: $y = 2x + 30$. Substitute: $x + (2x + 30) = 180$, so $3x + 30 = 180$, $3x = 150$ and $x = 50$, then $y = 130$. ($130^\circ$ is the larger angle; $20^\circ$ used complementary ($90^\circ$) instead of supplementary.)

Let correct $= c$ and incorrect $= i$: $c + i = 10$ and $5c - 2i = 36$. Substitute $i = 10 - c$: $5c - 2(10 - c) = 36$, so $5c - 20 + 2c = 36$, $7c = 56$ and $c = 8$. (2 is the incorrect count; 10 is the total number of questions.)

Clear the fractions. Multiply the first by 3: $x + 3y = 12$. Multiply the second by 2: $2x + y = 14$. Multiply $x + 3y = 12$ by 2: $2x + 6y = 24$, then subtract: $5y = 10$, so $y = 2$ and $x = 6$. ($x = 2,\; y = 6$ swapped the values; $x = 12,\; y = 14$ used the right-hand sides after clearing fractions.)

Subtract the equations to eliminate $y$: $(5x + 2y) - (3x + 2y) = 22 - 14$, so $2x = 8$ and $x = 4$. Substitute back: $3(4) + 2y = 14$, so $2y = 2$ and $y = 1$. Therefore $x - y = 4 - 1 = 3$. (5 gives $x + y$; the trap is solving correctly but giving the wrong expression.)

Let the width $= w$ and the length $= l$. Perimeter: $2(l + w) = 30$, so $l + w = 15$. Relationship: $l = 2w + 3$. Substitute: $(2w + 3) + w = 15$, so $3w + 3 = 15$, $3w = 12$ and $w = 4$ cm (the length is 11 cm). (11 cm is the length; 7.5 cm treated the rectangle as a square.)

Let the 10c coins $= x$ and the 20c coins $= y$: $x + y = 50$ and, in cents, $10x + 20y = 720$. Multiply the first by 10: $10x + 10y = 500$. Subtract: $10y = 220$, so $y = 22$ and $x = 28$. (28 is the 10c coin count; 50 is the total number of coins.)

Substitute each point into $y = mx + c$. From $(1, 5)$: $5 = m + c$. From $(3, 9)$: $9 = 3m + c$. Subtract: $2m = 4$, $m = 2$ and $c = 3$, so $m + c = 5$. Shortcut: substituting $(1, 5)$ gives $m + c = 5$ directly, so the answer is just 5. (4 gives $2m$; 2 gives just $m$.)