Simultaneous Equations (Algebraically)
Master simultaneous equations the algebraic way for NSW Year 12 Mathematics Standard 2. This topic shows how to solve a pair of linear equations by elimination and by substitution — finding the exact pair of values \((x,y)\) that satisfies both equations at once.
You will learn to match coefficients and add or subtract to eliminate a pronumeral, make a pronumeral the subject and substitute, back-substitute for the second value, and check your solution — a core Standard 2 skill that gives an exact answer where a graph can only be read off approximately.
Theory
Simultaneous linear equations are two straight-line equations solved together. This Year 12 Standard 2 (NSW) guide shows how to solve them algebraically — by elimination and by substitution — then back-substitute and check the pair \((x,y)\) that satisfies both equations exactly.
A pair of simultaneous linear equations is two straight-line equations that must both be true at the same time. Their solution is the single pair of values \((x,y)\) that satisfies both.
Solving algebraically means finding that pair by working with the equations themselves, rather than reading a graph. There are two standard methods: elimination and substitution. Both give the same exact answer.
Elimination combines the equations so one pronumeral cancels: match the coefficients of \(x\) or \(y\), then add or subtract. Substitution makes one pronumeral the subject of one equation and puts that expression into the other. Whichever you use, always check your pair in both original equations.
Write the pair in the general form (\(a,b,c\) are numbers):
Elimination. Multiply an equation so one pronumeral has matching coefficients, then add (opposite signs) or subtract (same signs) to remove it:
Substitution. Make one pronumeral the subject, then replace it in the other equation:
Elimination method
- Line up the equations so \(x\) terms, \(y\) terms and numbers match up.
- Match a coefficient: multiply one (or both) whole equations so \(x\) or \(y\) has the same size coefficient in each.
- Add or subtract the equations to eliminate that pronumeral, then solve for the one that is left.
- Back-substitute to find the other pronumeral, then check \((x,y)\) in both originals.
Substitution method
- Make a pronumeral the subject of one equation (e.g. \(x=\ldots\) or \(y=\ldots\)).
- Substitute that expression into the other equation, giving one equation in one pronumeral.
- Solve it, then back-substitute to find the other pronumeral.
- Check and state the solution \((x,y)\) in both original equations.
The \(y\) terms are \(+y\) and \(-y\), so adding the equations removes \(y\).
| \((4x+y)+(2x-y)\) | \(=\) | \(10+2\) |
| \(6x\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(2\) |
| \(2(2)-y\) | \(=\) | \(2\) |
| \(y\) | \(=\) | \(2\) |
| \(\therefore\ (x,y)\) | \(=\) | \((2,2)\) |
Check: \(4(2)+2=10\ \checkmark\).
Multiply \(x+y=6\) by \(2\) so both have \(2y\), then subtract.
| \(3x+2y\) | \(=\) | \(16\) |
| \(2x+2y\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(4\) |
| \(4+y\) | \(=\) | \(6\) |
| \(y\) | \(=\) | \(2\) |
| \(\therefore\ (x,y)\) | \(=\) | \((4,2)\) |
Check: \(3(4)+2(2)=16\ \checkmark\).
The first equation already gives \(y\), so replace \(y\) in the second.
| \(3x+(2x-3)\) | \(=\) | \(17\) |
| \(5x-3\) | \(=\) | \(17\) |
| \(5x\) | \(=\) | \(20\) |
| \(x\) | \(=\) | \(4\) |
| \(y\) | \(=\) | \(2(4)-3=5\) |
| \(\therefore\ (x,y)\) | \(=\) | \((4,5)\) |
Check: \(3(4)+5=17\ \checkmark\).
Rearrange the first equation to \(x=y+4\), then substitute.
| \(3(y+4)+4y\) | \(=\) | \(-2\) |
| \(7y+12\) | \(=\) | \(-2\) |
| \(7y\) | \(=\) | \(-14\) |
| \(y\) | \(=\) | \(-2\) |
| \(x\) | \(=\) | \(-2+4=2\) |
| \(\therefore\ (x,y)\) | \(=\) | \((2,-2)\) |
Check: \(3(2)+4(-2)=-2\ \checkmark\).
Common pitfalls
Frequently asked questions
How do you solve simultaneous equations algebraically?
Use elimination or substitution. Elimination combines the two equations so one pronumeral cancels; substitution puts one pronumeral in terms of the other and replaces it. Either way you solve for one value, back-substitute for the other, then check the pair in both equations.
When should I use elimination and when substitution?
Use substitution when one equation already has a pronumeral by itself, such as y equals 2x minus 3, because you can substitute straight away. Use elimination when the equations are both in the form ax plus by equals c, since lining them up and adding or subtracting is quicker.
Do you add or subtract the equations in elimination?
Match the coefficients of one pronumeral first. If the matching terms have opposite signs, add the equations; if they have the same sign, subtract. Both remove that pronumeral so you can solve for the other.
How do I check my solution?
Substitute your x and y values back into both original equations. If each equation balances, the pair is correct. Checking catches sign errors, which are the most common mistake.
What if the solution is a fraction or a negative number?
That is fine and common. Algebra gives the exact value even when it is negative or a fraction, which is the main advantage over reading a graph. Keep the sign and, for a fraction, leave it exact rather than rounding.
Is solving algebraically the same as solving graphically?
The answer is the same pair of values, which is where the two lines cross. Graphing shows the meaning but is only as accurate as your reading; elimination and substitution give the exact solution every time.