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Topos Uranos
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First-degree equations and inequalities, and linear systems

Summary

This lesson studies first-degree equations and inequalities and systems of linear equations. It proves which operations transform an equation into another with the same solutions and explains why multiplying by an expression that can vanish, or squaring, can add or remove solutions. It analyzes completely the equation ax+b=0ax + b = 0, also when its coefficients depend on a parameter, solves equations with fractions and linear inequalities, single and simultaneous, and expresses their solution sets as intervals. Finally, it solves systems of two equations in two unknowns by substitution, elimination and comparison, proves Cramer’s rule, classifies systems according to their determinant, solves a system of three equations by elimination and translates problems about mixtures, ages and speeds into systems.

Learning objectives

  1. Justify which operations preserve the solution set of an equation or of an inequality, and recognize those that can add or remove solutions.
  2. Analyze and solve first-degree equations, with parameters or with fractions, and linear inequalities, single and simultaneous, writing their solutions as intervals.
  3. Solve systems of two linear equations by substitution, elimination and comparison, and classify them according to their determinant by means of Cramer’s rule.
  4. Set up and solve linear systems that model problems, interpreting the cases of infinitely many solutions or of none.

Equations and equivalent equations

An equation in the unknown xx is an equality A=BA = B between two expressions that depend on xx. Its solutions are the real numbers x0x_0 for which both expressions are defined and take the same value; the set they form is the solution set of the equation. Solving an equation means describing that set, which may have one element, several, infinitely many or none. For example, the solution set of 2x−6=02x - 6 = 0 is {3}\{3\}, that of x+1=1+xx + 1 = 1 + x is the whole real line, and that of x+1=xx + 1 = x is empty.

Two equations are equivalent when they have the same solution set. The general method for solving an equation consists in replacing it by a succession of equivalent equations, each simpler than the previous one, until reaching one whose solution set can be read off directly, such as x=3x = 3. Now, each transformation in that succession must preserve the solutions, without adding or removing any; it follows that the first task is to know which operations guarantee this.

Operations that preserve the solutions

The two fundamental operations are adding the same expression to both sides and multiplying them by the same nonzero number. Subtracting and dividing are particular cases, because subtracting CC is adding −C-C, and dividing by kk is multiplying by k−1k^{-1}, as was seen in the lesson on number systems.

TheoremThe operations that preserve the solutions

Let AA, BB and CC be expressions in the unknown xx, defined on the same set of real numbers, and let kk be a nonzero real number. Then the equations A=BA = B, A+C=B+CA + C = B + C and kA=kBkA = kB have the same solution set. More generally, if the expression CC does not vanish at any point of that set, A=BA = B and CA=CBCA = CB have the same solution set. Two inclusions must be proved for each operation: every solution of the original equation is a solution of the transformed one, and conversely. The figure represents the equation as a balance in equilibrium, which keeps it if the same thing is done on both pans.

Proof

  1. A(x0)=B(x0)→A(x0)+C(x0)=B(x0)+C(x0)\dato{s}{A(x_0) = B(x_0)} \rightarrow A(x_0) + \resaltar{C(x_0)} = B(x_0) + \resaltar{C(x_0)}

    Let x0x_0 be a solution of A=BA = B: the numbers A(x0)A(x_0) and B(x0)B(x_0) are the same, and adding the same number to them gives the same result. Every solution of A=BA = B is a solution of A+C=B+CA + C = B + C.

  2. (A(x0)+C(x0))+(−C(x0))=A(x0)+0=A(x0)\left( A(x_0) + C(x_0) \right) + \resaltar{(-C(x_0))} = A(x_0) + 0 = \resaltar{A(x_0)}

    Conversely, if x0x_0 satisfies A+C=B+CA + C = B + C, we add to both sides the opposite of C(x0)C(x_0). By associativity and the definition of the opposite, only the first summand remains on each side, whence A(x0)=B(x0)A(x_0) = B(x_0): the two solution sets coincide.

  3. A(x0)=B(x0)→kA(x0)=kB(x0)A(x_0) = B(x_0) \rightarrow \resaltar{k} A(x_0) = \resaltar{k} B(x_0)

    In the same way, if x0x_0 is a solution of A=BA = B, multiplying the same number by kk gives the same result.

  4. k−1(kA(x0))=(k−1k)A(x0)=A(x0)\resaltar{k^{-1}} \left( k A(x_0) \right) = \left( k^{-1} k \right) A(x_0) = \resaltar{A(x_0)}

    Conversely, since k≠0k \neq 0, its inverse k−1k^{-1} exists; multiplying both sides of kA(x0)=kB(x0)kA(x_0) = kB(x_0) by it recovers the original equality. If CC does not vanish at any point, the same argument works at each x0x_0 with the number k=C(x0)k = C(x_0).

  5. 0⋅A(x0)=0=0⋅B(x0)\resaltar{0} \cdot A(x_0) = 0 = \resaltar{0} \cdot B(x_0)

    The hypothesis k≠0k \neq 0 is indispensable: when multiplying by 00, both sides equal 00 for every xx, and the transformed equation is satisfied by any number, whether or not it is a solution of the original one.

Operations that can add or remove solutions

If CC vanishes at some point, the equivalence can fail. When both sides are multiplied by an expression that vanishes at x1x_1, the number x1x_1 becomes a solution, even if it was not one; and when dividing by it, the solutions at which it vanishes are lost. For example, x=2x = 2 has a single solution, but after multiplying by x−3x - 3 the number 33 appears as well; and the equation x2=3xx^2 = 3x has the solutions 00 and 33, because x2−3x=x(x−3)x^2 - 3x = x(x - 3), but dividing by xx leaves x=3x = 3 and 00 is lost:

x(x−3)=2(x−3)↔x=2∨x=3,x2=3x↔x=0∨x=3x(x - 3) = 2(x - 3) \leftrightarrow x = 2 \vee x = 3, \qquad x^2 = 3x \leftrightarrow x = 0 \vee x = 3

Squaring both sides is not a reversible operation either, because two opposite numbers have the same square. The following lemma specifies which solutions can appear.

LemmaSquaring

For real numbers AA and BB, A2=B2A^2 = B^2 holds if and only if A=BA = B or A=−BA = -B. Consequently, squaring both sides of an equation A=BA = B preserves all its solutions, but the solutions of the equation A=−BA = -B may be added; if AA and BB take no negative values, none is added. The proof factors the difference of the squares. The figure draws the parabola y=t2y = t^2 and the horizontal line at height B2B^2, with B=2B = 2: the numbers whose square is B2B^2 are the abscissas of the two intersection points, BB and −B-B.

Proof

  1. A2−B2=(A−B)(A+B)A^2 - B^2 = \resaltar{(A - B)(A + B)}

    By distributivity, the difference of the squares factors; the terms ABAB and −AB-AB cancel.

  2. A2=B2↔(A−B)(A+B)=0↔A=B∨A=−B\dato{eq}{A^2 = B^2 \leftrightarrow (A - B)(A + B) = 0 \leftrightarrow A = B \vee \resaltar{A = -B}}

    A product is zero if and only if one of its factors is, as was proved in the lesson on number systems. Hence numbers with equal squares are equal or opposite.

  3. A2=B2↔(A=B)∨(A=−B)A^2 = B^2 \leftrightarrow \left( A = B \right) \vee \resaltar{\left( A = -B \right)}

    Applied at each value of xx, this says that the solutions of the squared equation are those of A=BA = B together with those of A=−BA = -B: none is lost, but those of the second may be added.

  4. A≥0≥−B=A→A=0=BA \geq 0 \geq -B = A \rightarrow \resaltar{A = 0 = B}

    If AA and BB are nonnegative and A=−BA = -B, the number AA lies between two equal quantities; whence A=0=BA = 0 = B, and the added solution was already a solution of A=BA = B.

Equations with roots, whose laws were studied in the lesson on powers with rational exponents and radicals, are solved by squaring; by the lemma, every solution of the resulting equation must be checked in the original one.

ExampleA solution that appears when squaring

Solve the equation x+2=x\sqrt{x + 2} = x. The figure shows the root, its opposite and the line y=xy = x: the squared equation collects the intersections of the line with both.

Proof

  1. (x+2)2=x2→x+2=x2\resaltar{\left( \sqrt{x + 2} \right)^2} = x^2 \rightarrow \resaltar{x + 2} = x^2

    The root is defined for x≥−2x \geq -2. We square both sides; the right-hand side can be negative, so the resulting equation may have more solutions than the original one.

  2. x2−x−2=(x−2)(x+1)=0\resaltar{x^2 - x - 2} = \resaltar{(x - 2)(x + 1)} = 0

    We move everything to the right-hand side and factor; the product (x−2)(x+1)(x - 2)(x + 1), when expanded, gives x2−x−2x^2 - x - 2.

  3. x=2∨x=−1\dato{c}{x = 2 \vee x = -1}

    A product is zero if and only if one of its factors is: there are two candidates.

  4. 2+2=4=2\sqrt{\resaltar{2} + 2} = \sqrt{4} = \resaltar{2}

    We check the first one in the original equation: it is a solution.

  5. −1+2=1≠−1\sqrt{\resaltar{-1} + 2} = 1 \neq \resaltar{-1}

    The second one is not: the left-hand side equals 11 and the right-hand side −1-1. It is a solution of −x+2=x-\sqrt{x + 2} = x, the equation A=−BA = -B that squaring added. Therefore, the solution set is {2}\{2\}.