Mathematics · Expressions, equations and inequalities · Grade 7

Systems of equations

2 kg of apples and 3 kg of pears — 29 000; 1 kg and 1 kg — 12 000 soum. The prices?

The situation

At the market, 2 kg of apples and 3 kg of pears cost 29 000 soum, and 1 kg of apples and 1 kg of pears cost 12 000 soum. How much do apples and pears cost separately? There are two unknowns, but also two conditions. Each condition is a straight line in the coordinate plane, and the pair that satisfies both is their point of intersection. We find it first by drawing, then exactly by calculation.

Topic

Systems of linear equations: the graph, substitution and elimination, problems. At the market 2 kg of apples and 3 kg of pears cost 29 000 soum, 1 kg and 1 kg cost 12 000 soum: the point where two straight lines meet is the solution

Key ideas

Graphical method
Substitution
Elimination

Formulas

{\begin{matrix} a_{1} x + b_{1} y = c_{1} \\ a_{2} x + b_{2} y = c_{2} \end{matrix}

A system of two linear equations in two unknowns. A solution is a pair $(x; y)$ that makes both equations true.

y = 12 - x \Rightarrow 2 x + 3 (12 - x) = 29

Substitution: we express $y$ from one equation and substitute it into the other — one unknown is left.

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow no solution

If the coefficients are proportional, the lines are parallel; if the constant terms are proportional too, the lines coincide (infinitely many solutions).

(2 x + 3 y) - 3 (x + y) = 29 - 36 \Rightarrow - x = - 7

Elimination (addition) method: we multiply the equations so that when we add (subtract) them, one unknown disappears.

Theory

Two unknowns, one equation

The solutions of the equation $x + y = 12$ are pairs $(x; y)$ : $(0; 12)$ , $(1; 11)$ , $(7; 5)$ , $(2.5; 9.5)$ … — infinitely many. In the coordinate plane they all lie on one straight line: the graph of the linear function $y = 12 - x$ . On the first step, drag the point along the line — the sum is always 12.

A system and the graphical method

The second condition $2 x + 3 y = 29$ is a second straight line. A pair that satisfies both equations lies on both lines, that is, at their point of intersection: $(7; 5)$ . There are three cases: the lines intersect — one solution; they are parallel — no solution; they coincide — infinitely many. Parallelism can be seen from the coefficients: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$ . The graphical method shows the answer, but fractional coordinates cannot be read exactly from a drawing — that is why algebraic methods are needed.

The substitution method

From one equation we express one of the unknowns: from $x + y = 12$ , $y = 12 - x$ . We substitute this into the second equation: $2 x + 3 (12 - x) = 29$ — now there is one unknown, an ordinary linear equation: $2 x + 36 - 3 x = 29$ , $x = 7$ . Then $y = 12 - 7 = 5$ . Geometric meaning: the point moves along the first line ( $y$ is always $12 - x$ ) and stops when it reaches the second line.

Elimination

We multiply the equations by numbers so that the coefficients of one unknown are equal (or opposite): multiplying $x + y = 12$ by 3 gives $3 x + 3 y = 36$ . Now we subtract it from the second equation: $(2 x + 3 y) - (3 x + 3 y) = 29 - 36$ , that is, $- x = - 7$ , $x = 7$ . Any such combination of the equations is again a straight line, and it too passes through the solution point: on the fourth step the pencil of lines shows this.

Solving word problems with a system

We call the two unknown quantities $x$ and $y$ and form two equations from the two conditions of the problem. Motion towards each other: two cars covered a distance of 300 km in 2 hours — $2 x + 2 y = 300$ ; the difference of speeds is 10 km/h — $x - y = 10$ . Solution: $x = 80$ , $y = 70$ km/h. We check the answer against the problem: $160 + 140 = 300$ . Price, mixture and work problems go the same way.

About the model

In the drawing the axes run from $- 1$ to $13.5$ : an intersection point outside this window is not visible, but the calculation still finds it. The graphical method is approximate: reading the coordinates of a point from a drawing can introduce an error, while substitution or elimination gives the exact answer. The experiment deals only with linear systems in two unknowns; systems with three or more unknowns are solved by the same methods, but in more steps.

Did you know?

The idea of representing an equation by a line in the coordinate plane comes from René Descartes's “Geometry”, published in 1637 — that is why rectangular coordinates are also called “Cartesian coordinates”.
In the ancient Chinese “The Nine Chapters on the Mathematical Art” (compiled no later than the 1st century AD), the eighth chapter solves systems of linear equations in a way similar to today's Gaussian elimination.

Inside the experiment

The topic is explained step by step with animation, from simple to advanced: from a real-life picture to the formula and the proof. Then you drag the points yourself to check. Inside are missions, a quiz and a guided tour.

More experiments in this section

Expressions: from numbers to letters — Grade 7
Monomials and polynomials — Grade 7
Special products (algebraic identities) — Grade 7
Algebraic fractions — Grade 7
The linear equation: the balance — Grade 7
Inequalities: ray and interval — Grade 8
The quadratic equation: the parabola — Grade 8