How to Solve a System of Equations: Graphing, Substitution, Elimination

The race, in symbols: hiker $d=2t$ kilometers after $t$ hours; late-starting cyclist $d=6(t-1)$. One line cannot name the meeting moment; two lines together can - that pairing is the setup for how to solve a system of equations.

The lines cross at $t=1.5$, $d=3$: an hour and a half in, both stand 3 km out. The picture shows how to solve a system of equations in its gentlest form. The algebra below does it ruler-free - three ways, each a different key of how to solve a system of equations.

A system of linear equations, by definition: two or more linear equations working together as one set form a system - the brace in $\{d=2t,\ d=6(t-1)\}$ only marks the grouping. How to solve a system of equations starts there.

A solution of a system of equations is the pair true in every member at once. Test $(1.5,\,3)$: $2\times1.5=3$ holds, $6\times(1.5-1)=3$ holds. True twice - the pair every method of how to solve a system of equations aims at. One equation never pins it: Ray's 1852 Algebra notes a value "can become known only when the values of the rest are given." One line holds infinitely many pairs; the second strands one - that squeeze is how to solve a system of equations entire.

Road one: how to solve a system of equations by graphing. Draw each equation as a line, find the intersection, check it in both equations.

Worked example. Solve $x-3y=-4$ and $x+y=4$. Slope form: $y=\frac{x+4}{3}$ climbs from $(-4,0)$; $y=4-x$ falls through $(0,4)$ - they cross at $(2,2)$.

Check: $2-3\times2=-4$ and $2+2=4$ - both true: $(x,y)=(2,2)$, the whole solution.

Solve by graphing: the three endings

Graphing also shows the three endings: crossing lines, one solution; parallel lines, none - inconsistent; one line twice, infinitely many. Whole-number answers flatter this road. Fractional, off-grid crossings send it to algebra: the next two roads run how to solve a system of equations in pure symbols.

Road two: how to solve a system of equations by substitution - one move, a swap. Solve one equation for a variable, drop the expression into the other equation, and one variable remains. That simplicity makes substitution the first route into how to solve a system of equations.

Worked example. Solve $x+3y=10$ and $2x+y=5$. The first solves cleanly: $x=10-3y$. Into the second: $2(10-3y)+y=5$, so $20-5y=5$, so $y=3$. Back-substitute: $x=10-3\times3=1$. Check: $1+9=10$, $2+3=5$ - solution $(1,3)$.

Ray's 1852 calls this elimination by substitution: find one unknown in one equation, use it in the other. Reach for it when an equation comes pre-solved - a naked $y=3x-2$ is an open invitation. Swap, solve, return, check - often the fastest way to solve for x and y.

So far you have seen how to solve a system of equations by graphing and by substitution; elimination is the third road, and the most mechanical. Ray's 1852: elimination is "deducing, from two or more equations containing two or more unknown quantities, a single equation containing only one unknown quantity." Today: combine 2 equations so one variable cancels.

Worked example. Stack $2x+y=7$ and $3x-y=13$. The $y$-terms, $+y$ and $-y$, are exact opposites. Add straight down: $5x=20$, so $x=4$. Back-substitute: $2\times4+y=7$, so $y=-1$. Solution $(4,-1)$; check: $3\times4-(-1)=13$.

This is adding down: two equations in, one variable out. With ready opposites in standard form, how to solve a system of equations rarely gets cleaner: stack, combine 2 equations, finish one-variable. This adding road is how to solve a system of equations by elimination before any scaling enters.

Sometimes nothing cancels - then scale first, and how to solve a system of equations gains its sharpest tool. Ray's rule: "Multiply, or divide the equations, if necessary, so that one of the unknown quantities will have the same coefficient in both." Then add or subtract so the matched terms cancel.

Worked example - Ray's own 1852 exercise: $2x+3y=33$ and $4x+5y=59$. Multiply the first by 2, every term: $4x+6y=66$. Subtract the second: $(4x+6y)-(4x+5y)=66-59$, so $y=7$. Back-substitute: $2x+21=33$, so $x=6$ - solution $(6,7)$. From there, how to solve a system of equations by elimination runs exactly as in the last section: combine, back-substitute, check.

That is how to do elimination in algebra when coefficients refuse to match: scale, then combine - the whole equation multiplied, or the cancellation is fake. Signs alike, subtract; opposite, add.

Solve the system of equations using elimination, step by step

Stack both equations in standard form. Aim one variable's coefficients at each other, multiplying as needed. Add or subtract straight down, back-substitute the winner, then check the pair in both originals - that check signs how to solve a system of equations by elimination off.

A field guide turns how to solve a system of equations into three if-then rules: need a picture - graph; one equation pre-solved - substitute; both in standard form - eliminate.

How to find x and y from two equations? Read before you compute. A naked $y=3x-2$ begs for substitution; opposite or matching coefficients beg for elimination; whole-number answers flatter graphing. Most systems of equations in Algebra 1 are two lines meeting once. The table sums up how to solve a system of equations three ways:

Then take the road with the fewest fractions.

Whichever road answers how to solve a system of equations, it ends at one pair that solves for x and y together and checks in both.

1. Trusting the graph too far. $(1.5,\,3)$ reads cleanly; $(\frac{4}{3},\,\frac{11}{5})$ does not - graphing is imprecise off integers. See it by graphing, get it exactly by algebra - the core split of how to solve a system of equations.

2. Substituting into the same equation. From $x+3y=10$, feeding $x=10-3y$ into its own line gives $10-3y+3y=10$ - true, useless. The expression goes into the other equation - how to solve a system of equations needs both talking.

3. Wrong add/subtract, or partial multiply. Scaling $2x+3y=33$ by 2 gives $4x+6y=66$ only if every term rides. Ray's rule: signs alike, subtract; opposite, add. A partial multiply fakes the cancellation - and how to solve a system of equations dies quietly there.

4. Forgetting the back-substitution. $x=4$ is half an answer; return for $y=-1$ and report $(4,-1)$.

5. Skipping the double check. Two true substitutions close the case; unchecked, the pair is a rumor.