Solving Simultaneous Equations

The values \( x = 5 \), \( y = 1 \) do satisfy the first equation — \( 2(5) + 1 = 11 \) ✓ — but they do not satisfy the second: \( 5 + 1 = 6 \), not 7. A solution to simultaneous equations must satisfy both equations at the same time. Satisfying only one equation is not enough.

The correct solution is found by subtracting: \( (2x + y) \; – \; (x + y) = 11 \; – \; 7 \), giving \( x = 4 \). Then \( 4 + y = 7 \), so \( y = 3 \). Check: \( 2(4) + 3 = 11 \) ✓ and \( 4 + 3 = 7 \) ✓. Both equations are satisfied simultaneously.

Both equations say “\( x + y \) equals something”, but they claim \( x + y = 10 \) and \( x + y = 6 \) at the same time. This would require \( 10 = 6 \), which is impossible. No pair of values for \( x \) and \( y \) can make the same expression equal two different numbers simultaneously.

Graphically, \( x + y = 10 \) and \( x + y = 6 \) are parallel lines (both have gradient \( -1 \)) with different \( y \)-intercepts. Since they never meet, there is no point \( (x, y) \) that lies on both lines.

Both equations have \( +2y \). When we subtract: \( (3x + 2y) \; – \; (x + 2y) = 19 \; – \; 11 \), giving \( 2x = 8 \), so \( x = 4 \). The \( y \)-terms cancel because \( 2y \; – \; 2y = 0 \). We can then find \( y \): \( 4 + 2y = 11 \), so \( y = 3.5 \). Check: \( 3(4) + 2(3.5) = 19 \) ✓ and \( 4 + 2(3.5) = 11 \) ✓.

When the coefficients of the variable you want to eliminate have the same sign, you subtract. If instead you added: \( 4x + 4y = 30 \) — neither variable is eliminated, and you still have two unknowns. Subtraction works because same-sign terms cancel under subtraction; opposite-sign terms cancel under addition.

Multiplying gives \( 2x + 6y = 28 \). If \( x + 3y = 14 \) is true for some pair \( (x, y) \), then doubling both sides preserves the equality: \( 2(x + 3y) = 2(14) \). The left side expands to \( 2x + 6y \) and the right to 28. Any pair satisfying the original also satisfies the new equation, and vice versa (dividing by 2 reverses it).

The key is that every term must be multiplied — both the \( x \), the \( 3y \), and the 14. A common error is to multiply only the term you’re focusing on (e.g. writing \( 2x + 3y = 14 \)), which changes the solutions entirely. Multiplying a complete equation by a non-zero constant produces an equivalent equation.

Example: \( 3x + y = 10 \) and \( 3x + 2y = 14 \). The \( x \)-coefficients are both 3, so subtract to eliminate \( x \): \( -y = -4 \), giving \( y = 4 \), \( x = 2 \).

Another: \( 5x + y = 16 \) and \( 5x \; – \; 2y = 1 \). Subtract: \( 3y = 15 \), so \( y = 5 \), \( x = 2.2 \).

Creative: \( x + 4y = 17 \) and \( x \; – \; y = 2 \). The \( x \)-coefficients are both just 1 — students often forget that “1” counts as a matching coefficient. Subtract: \( 5y = 15 \), \( y = 3 \), \( x = 5 \).

Trap: \( x + 2y = 8 \) and \( 3x + y = 9 \). The \( x \)-coefficients are 1 and 3, which are not equal, so subtracting does not eliminate \( x \). Subtracting gives \( -2x + y = -1 \), which still has both unknowns. A student might think “you always just subtract” without checking that the coefficients match.

Example: \( x + y = 5 \) and \( x + y = 8 \). The same expression cannot equal two different values.

Another: \( 2x + 3y = 10 \) and \( 2x + 3y = 4 \). Same left-hand side, different right-hand sides.

Creative: \( 3x \; – \; y = 7 \) and \( 6x \; – \; 2y = 20 \). These don’t look obviously incompatible, but the second is almost double the first: \( 2(3x \; – \; y) = 14 \neq 20 \). The left-hand sides are proportional but the right-hand sides are not, so the lines are parallel.

Trap: \( x + y = 5 \) and \( 2x + 2y = 10 \). A student might offer this thinking the equations look different. But the second is simply \( 2 \times \) the first — they represent the same line and have infinitely many solutions, not zero solutions.

Example: \( 2x + 3y = 12 \) and \( x + y = 5 \). No coefficients match directly. Multiply the second by 2: \( 2x + 2y = 10 \). Subtract: \( y = 2 \), \( x = 3 \).

Another: \( 3x + y = 14 \) and \( x + 2y = 8 \). Multiply the first by 2: \( 6x + 2y = 28 \). Subtract: \( 5x = 20 \), \( x = 4 \), \( y = 2 \).

Creative: \( 2x + 5y = 24 \) and \( 3x + 4y = 22 \). Neither coefficient matches, so you must multiply both equations. Multiply first by 3 and second by 2: \( 6x + 15y = 72 \) and \( 6x + 8y = 44 \). Subtract: \( 7y = 28 \), \( y = 4 \), \( x = 2 \).

Trap: \( 2x + 5y = 10 \) and \( 2x + 3y = 6 \). A student might offer this, focusing on the \( 5y \) and \( 3y \) which are different, and assume multiplication is needed. But looking closer, the \( 2x \) terms are already identical! You can subtract immediately without multiplying anything.

Example: \( x + y = 2 \) and \( x \; – \; y = -4 \). Add: \( 2x = -2 \), \( x = -1 \), \( y = 3 \).

Another: \( 2x + y = -1 \) and \( x + y = 1 \). Subtract: \( x = -2 \), \( y = 3 \).

Creative: \( x + 2y = 3 \) and \( 2x + y = 0 \). From the second: \( y = -2x \). Substituting: \( x + 2(-2x) = 3 \), so \( -3x = 3 \), \( x = -1 \), \( y = 2 \). Neither equation has a negative constant, yet the solution still has \( x < 0 \).

Trap: \( 2x + y = 5 \) and \( x \; – \; y = 1 \). A student might think the minus sign in \( x \; – \; y \) means \( x \) must be negative. But adding: \( 3x = 6 \), \( x = 2 \), \( y = 1 \) — both values are positive. The minus sign in the equation is a subtraction operation, not a statement about the sign of the solution.

Most pairs of linear simultaneous equations do have a unique solution — this happens whenever the lines intersect at a single point. For example, \( x + y = 7 \) and \( x \; – \; y = 3 \) gives \( x = 5, y = 2 \) — one solution.

However, there are two exceptions. If the lines are parallel (same gradient, different intercept), there is no solution at all — e.g. \( x + y = 5 \) and \( x + y = 3 \). If the equations represent the same line (one is a multiple of the other), there are infinitely many solutions — e.g. \( x + y = 5 \) and \( 2x + 2y = 10 \). Students often assume “simultaneous equations” always means “one answer”.

If the coefficients are identical (same value and same sign), subtracting always eliminates that variable. For example, in \( 3x + 2y = 16 \) and \( x + 2y = 10 \), the \( y \)-coefficients are both \( +2 \). Subtracting: \( 2y \; – \; 2y = 0 \), eliminating \( y \).

A helpful memory hook is SSS: Same Sign Subtract. If coefficients are identical (e.g. both \( +2y \)), subtract. If they are opposite signs (e.g. \( +2y \) and \( -2y \)), add.

Many textbook problems are designed so the answers are whole numbers, which can lead students to believe this is always the case. For example, \( x + y = 10 \) and \( x \; – \; y = 4 \) gives the whole-number solution \( x = 7, y = 3 \).

But consider \( x + 2y = 7 \) and \( 3x + y = 8 \). Multiplying the first by 3: \( 3x + 6y = 21 \). Subtracting: \( 5y = 13 \), so \( y = 2.6 \) and \( x = 1.8 \). Solutions can be any rational numbers (fractions or decimals), not just whole numbers. Students who expect “nice” answers may assume they’ve made an error when they get a non-integer — but it may well be correct.

This is the fundamental principle that makes the elimination method valid. If \( a = b \) and \( c = d \), then \( a + c = b + d \) — adding equal quantities to equal quantities preserves equality. For example, if \( 2x + y = 7 \) and \( x \; – \; y = 2 \), then \( (2x + y) + (x \; – \; y) = 7 + 2 \), giving \( 3x = 9 \).

Students sometimes apply elimination mechanically without understanding why it works. This can lead to errors when they’re unsure whether to add or subtract, or when they apply the operation to only one side. Understanding that adding (or subtracting) entire true equations produces a new true equation is the logical foundation of the elimination method.

The student only multiplied the \( x \)-term by 2, writing \( 2x + 3y = 11 \) instead of the correct \( 2x + 6y = 22 \). This is the “not multiplying all terms” misconception. When you multiply an equation by a number, every term must be multiplied — both the variable terms and the constant. The correct scaling is \( 2(x + 3y) = 2(11) \), giving \( 2x + 6y = 22 \).

With the correctly scaled equation, subtracting gives \( (2x + y) \; – \; (2x + 6y) = 7 \; – \; 22 \), so \( -5y = -15 \), \( y = 3 \), and \( x = 2 \). Check: \( 4 + 3 = 7 \) ✓ and \( 2 + 9 = 11 \) ✓. The student’s answer of \( x = 2.5, y = 2 \) fails the check: \( 2.5 + 6 = 8.5 \neq 11 \).

The answer \( x = 3, y = 2 \) is indeed correct! However, the method is risky. Trial and improvement (“guess and check”) works here because the numbers are small and simple.

For a system like \( 7x + 3y = 29 \) and \( 2x \; – \; 5y = -8 \), or if the answers were decimals like \( x = 3.4 \), guessing would be incredibly difficult and time-consuming. Algebraic methods (elimination or substitution) are “robust” — they work for every pair of equations, not just the nice ones.

The student tried to use elimination without aligning the equations first. In the second equation, the \( y \) is on the left and the \( x \) is on the right, which doesn’t match the first equation’s layout.

To use elimination, you must rearrange the equations so variables line up (e.g., \( -x + y = 4 \)). Alternatively, since the second equation already tells you \( y = \dots \), this is a perfect candidate for substitution: replace the \( y \) in the first equation with \( (x + 4) \).

The elimination step is perfectly correct — \( x = 3 \). But the student stopped too early. This is the “forgetting the second variable” misconception. A solution to simultaneous equations requires values for both variables, not just one. The student must substitute \( x = 3 \) back into either equation to find \( y \).

Using \( x + y = 4 \): \( 3 + y = 4 \), so \( y = 1 \). The complete solution is \( x = 3, y = 1 \). Check: \( 3(3) + 1 = 10 \) ✓ and \( 3 + 1 = 4 \) ✓. Writing only “\( x = 3 \)” would lose marks in an exam — both values are needed for a complete solution.