Linear equations (With One Unknown)

Linear Mathematics & Matrices, pt.1 Dr. H. Schellinx

Linear equations A bit of terminology… In order to be able to apply mathematical techniques to problem solving we need to translate problems into mathematical language, which uses algebraic expressions. An algebraic expression is a combination of variables and numbers by means of the usual algebraic operations of addition, multiplication, exponentiation and their inverses. Variables (also called the unknowns) are letters (like x or y or t or v…) that represent some unknown quantity. Addition symbols chop algebraic expressions into parts that we call terms.

Example -‐ a+ b − 7x (= a+ b + −

7x

"

#$

%

&' ) is an algebraic expression. It is a combination

of three terms: a , b and − 7x. The variables are a, b and x.

A term without variables is called a constant. Two algebraic expressions that are joined by an equality sign (=) form an (algebraic) equation.

Example -‐ t2 + v3= z− 2 is an equation. It consists of the two algebraic expressions

t2 + v3and z− 2 . The algebraic expression to the left of the equality sign is a term,

which is the square root of yet another algebraic expression ( t2 + v3), which in turn is a

combination of the terms t2 and v3. The algebraic expression to the right of the equality

sign is a combination of the terms z and –2. The second of these terms, –2, is a constant. Linear equations… A linear equation is an algebraic equation in which each term is either a constant (a number like 3, or –6.7, or ½, or 𝑒!!, or 𝜋, or 7, or … etc. …) or the product of a number and (the first power of) a single variable (like –5x, or 1020z, or ½a, or !

!𝜋𝑏, or 𝑤 177, or … etc. ….)


Example -‐ The following are examples of linear equations: 3x + 5y+ z = 6

−7x + 72y = z

x 7 + 29y = 17 The following are examples of algebraic equations that are not linear equations: 9 = x2 + y2 19− x.y =123 y =17 2x +w

Exercise 1 -‐ Explain why the first three examples are linear equations, and why the last three examples are not. For each of the equations, indicate which are the constants.

A solution of an algebraic equation consists in a series of numbers that, when substituted for the variables in the equation, give rise to a true statement: the result of the substitution is an identity. To solve an equation means: find all possible solutions, i.e. find all the numbers that, when substituted for the variables in the equation, give rise to a true statement. Example -‐ The ordered pair or tuple (x,y) = (1,1) [i.e. the numbers x = 1 and y = 1] is a solution of the linear equation x + 4y = 5. But it is not the only solution. For example, also the tuples (5,0) and (0, 1.25) are solutions.

Exercise 2 -‐ Can you find all solutions (we will often say: the solution) of the linear equation x + 4y = 5? Exercise 3 -‐ For the following algebraic expressions: find the terms, the variables and the constants. If the expression is an equation, is it a linear equation? If it is not, explain why. Can you solve the equations?

a. 33𝑥 + 42 𝑦

b.

c.

d.

e. f. 𝑥 − 3! = 𝑦 + 1

g.

Exercise 4 -‐ Is x −1( )2 = 4 a linear equation? Can you solve it?

110

x + z2

16+ 7

a+ 3b=1

8p+ sin(π ) = 0a− b

cos(x)−3y = t2


A linear equation with n different variables, say x1, x2, x3,..., xn , can always be written in the form a1x1 + a2x2 + a3x3 +...+ anxn = c , where the ai are numbers; they are called the coefficients of the equation. Also c is a number. It is the constant of the equation. A linear equation of the form is said to be in standard form.

Fact: Every linear equation can be transformed into an equivalent equation in standard form.

Example -‐ The equation 7x −8+ 2y+3z = y+ x − 2 is a linear equation with 3 unknowns (x, y and z), but it is not in standard form. We transform it step by step, into an equivalent equation that is in standard form, as follows: 7x −8+ 2y+3z = y+ x − 2 ⇔ (7x − x)−8+ 2y+3z = y− 2 ⇔ (7x − x)−8+ (2y− y)+3z = −2 ⇔ (7x − x)+ (2y− y)+3z = −2+8 ⇔ 6x + y+3z = 6 5x −3y = z+w− 6 is a linear equation with 4 unknowns (x,y,w and z). It has the equivalent standard form 5x −3y−w− z = −6 . −3y = y+ 2x − 7 is a linear equation with 2 unknowns (x and y). It has the equivalent standard form 2x + 4y = 7 . NB.: We will always put the unknowns in a standard form in alphabetical order, and, for example, not write b + 2a = 14, but: 2a + b = 14.

Exercise 5 -‐ Write each of the following linear equations in standard form. Then solve the equations.

a. 3(b+ 2)− 2 = −(5+ b)+ b

b.8x +3(2− x) = 5x + 6

c. 35x + 0.7 = x − 4

5

d. 2x − 6 = −2x + 4(x − 2)

e. 0.06(a+ 200)+ 0.1a =172 Exercise 6 -‐ Translate the following problem into mathematical language, by writing it as two linear equations with two unknowns, c and t. Such a pair of equations is also called ‘a system of linear equations’. Can you solve this system? “A student has a car loan (c) charging 0.75% interest per month and a tuition loan (t) charging 0.5% interest per month. How much does she owe on each account if she pays a total of € 395.50 monthly interest on a total debt of € 62,200?”

a1x1 + a2x2 + a3x3 +...+ anxn = c


Linear equations with one unknown Linear equations with only one variable are the simplest ones. We can always write such an equation in the standard form ax = b . Here x is the only variable (the unknown). The coefficient a and the constant b are both numbers.

1. If a ≠ 0 , the equation has precisely one solution: x = ba.

2. If a = 0 and b ≠ 0 , the equation has no solution. It is a contradiction. 3. If a = 0 and b = 0 , all real numbers satisfy the equation. It is an identity.

Recall that the root or zero of a function f(x) is a member x of the domain of f such that f(x) = 0 (we say that f(x) vanishes at x). The solution of the linear equation with one unknown ax = b is the root of the linear function f (x) = ax − b . In the two dimensional ‘Euclidean plane’, using a rectangular coordinate system, the graph of this function is the straight line represented by the equation y = ax − b . The coefficient a is the slope of the line. It indicates how steep the line is. If the slope is positive (a > 0), you will see that the line is going up when you move along the horizontal (x-‐)axis from the left to the right (from West to East). If the slope is negative (a < 0), you will see the line is going down when you move along the x-‐axis from the left to the right (from West to East). The constant b is called the line’s y-‐intercept: it tells us where the line crosses the vertical (y-‐)axis. The solution of the equation is the point where the line crosses the x-‐axis of our rectangular coordinate system: it is the x-‐intercept. As an example, look at the following linear equation with one unknown: 2x = 4 . The corresponding linear function is f (x) = 2x − 4 , which has as its graph the line y = 2x − 4 . The solution of the equation is the zero of the function, i.e. the point where the line crosses the x-‐axis: (2,0).


There are many kinds of often recurring practical problems that are most easily solved by interpreting them in the form of a linear equation with one unknown. Here are a few examples. You may remember them from the ‘Intermediate Algebra’ course: Problem 1. (Uniform Motion Problem) A cyclist leaves his training base for a morning workout, riding at the rate of 18 kilometers per hour. One hour later, his support staff leaves the base in a car, going 45 kilometers per hour in the same direction. How long will it take the support staff to catch up with the cyclist? Solution: Suppose that it takes the staff x hours to catch up with the cyclist. As their car’s speed is 45 km/h, the distance travelled by the support staff in x hours equals 45x kilometres. The cyclist left one hour earlier. When the staff’s car catches up with him, he has travelled a distance of 18(x+1) kilometres. The two distances have to be equal, so the linear equation that we need to solve is 45x =18(x +1) , which is equivalent to the standard form 27x =18 . We therefore find that the support staff catches up with the cyclist after 18/27 hours = 2/3 hours = 40 minutes. Problem 2. (Simple Investment Problem) An investment club invested part of €100.000 at 9% annual interest, and the rest at 8%. If the annual income from these investments was €8.600, how much was invested at 8%? Solution: Suppose that the investment club invested x euros at 8% annual interest. Then the club invested 100.000 – x euros at 9% annual interest. The linear equation that we need to solve is0,08x + 0,09(100.000− x) = 8.600 , which is equivalent to the standard form 0,01x = 400 . We therefore find that the investment club invested €40.000 at 8%. Problem 3. (Percent Mixture Problem) A dairy mixes milk with 4% of butterfat and milk with 1% of butterfat to get milk with 2% of butterfat. How much of the 4% butterfat milk does the dairy need to mix with how much of the 1% butterfat milk to produce 120 liters of milk with 2% of butterfat? Solution: Suppose the dairy needs x liters of 4% butterfat milk to produce 120 liters of 2% butterfat milk. Then it needs 120 – x liters of 1% butterfat milk. The amount of butterfat in the 2% mix has to be equal to the sum of the amounts in the parts: the linear equation that we need to solve is 0,04x + 0,01(120− x) = 0,02 ⋅120 , which is equivalent to the standard form 0,03x =1,2 . Thus we find that the dairy needs to mix 40 liters of 4% butterfat milk with 80 liters of 1% butterfat milk in order to obtain 120 liters of 2% butterfat milk.

Exercise 7 – What are the y-‐intercept b and the slope a of the line that connects the two points (-‐1, 3) and (2,0)? Exercise 8 (*) – Find expressions for the y-‐intercept b and the slope a of the line that connects the two points (x1, y1) and (x2, y2) in terms of the coordinates x1, y1, x2 and y2 of the two points.

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Linear equations (With One Unknown)