Two unknowns: when we need a system of equations

Two notebooks and one pen cost 110 hryvnias; one notebook and two pens cost 100. What is each price? A notebook costs 40 hryvnias and a pen 30. This pair must fit both receipts simultaneously: satisfying just one is not enough.

From one equation and word-problem models, we move to two linked conditions. Let $x$ be the notebook price and $y$ the pen price in hryvnias. Prices are unchanged between purchases, with no extra fees or discounts. This model treats prices as real numbers without rounding to kopecks.

A solution is a pair, not one number

We obtain a system of equations, conditions that must hold together:

\begin{cases}2x+y=110,\\x+2y=100.\end{cases}

The pair $(x,y)=(40,30)$ gives $80+30=110$ and $40+60=100$ . Order matters: $(30,40)$ gives 100 on the first receipt and fails.

The first equation alone admits many pairs: $(40,30)$ , $(30,50)$ , $(50,10)$ . The general form $ax+by=c$ , with $a,b$ not both zero, is a linear equation in two variables. A second independent relationship can select one pair.

See the shared point on a graph

Plot $x$ horizontally and $y$ vertically. The point $(40,30)$ records these two prices. All pairs satisfying the first receipt form the line $y=110-2x$ ; the second gives $y=(100-x)/2$ . Their intersection belongs to both lines and therefore fits both receipts.

One solution

x + 2y = 100

None

4x + 2y = 240

Infinitely many

4x + 2y = 220

Both axes are prices in hryvnias (UAH), with matching scales. Solid line: 2x+y=110. The dashed line represents the second equation printed below each panel. Only nonnegative values are shown; the lines continue.

A drawing suggests an answer, but coordinates read from a screen may be approximate. Algebra and substitution establish an exact result. Systems and solution methods, OpenStax §11.1.

Substitution leaves one unknown

From the first equation, $y=110-2x$ . Substitute this whole expression for $y$ in the second:

x+2(110-2x)=100.

Expanding gives $-3x=-120$ , so $x=40$ . Return to the first: $y=110-80=30$ . Returning matters: $x$ alone is not yet a solution pair.

Addition eliminates one variable

Alternatively, double the first equation: $4x+2y=220$ . Subtract the second, $x+2y=100$ , to obtain $3x=120$ . Again $x=40$ , $y=30$ .

We keep one original equation together with the new one. Replacing the whole system by its difference is invalid: $3x=120$ says nothing about $y$ . With one original condition retained, the transformation reverses and preserves the common solutions.

Two receipts need not be independent

If the second receipt is $4x+2y=220$ , it merely doubles the first. The lines coincide, giving infinitely many pairs described by $y=110-2x$ . Nonnegative prices restrict this to $0\le x\le55$ , still infinitely many possibilities in this real-number model.

If the second instead says $4x+2y=240$ , it contradicts the first: doubling the same purchase should cost 220. Subtraction yields $0=20$ ; the lines are parallel and share no solution. Under our fixed-price assumptions, the data are inconsistent, rather than awaiting an undiscovered price.

Branch · An edge caseClarifying the condition on coefficients.

When an equation does not describe a line

$0x+0y=0$ allows every pair: the whole plane. $0x+0y=5$ allows none. This is why the claim that a linear equation graphs as a line requires at least one nonzero variable coefficient.

A system seeks a common answer to every condition. Substitution and elimination find it exactly; a graph explains uniqueness, absence or ambiguity. Check the final pair in every original equation.