# Chapter 2. "Of the Resolution of Simple Equations, or Equations of the First Degree."

### Part I. Section IV. Chapter 2. “Of the Resolution of Simple Equations, or Equations of the First Degree.”

573 When the number sought, or the unknown quantity, is represented by the letter \(x\), and the equation we have obtained is such, that one side contains only that \(x\), and the other simply a known number, as, for example, \(x=25\), the value of \(x\) is already known. We must always endeavour, therefore, to arrive at such a form, however complicated the equation may be when first obtained: and, in the course of this section, the rules shall be given, and explained, which serve to facilitate these reductions.

574 Let us begin with the simplest cases, and suppose, first, that we have arrived at the equation \(x + 9 = 16\). Here we see immediately that \(x=7\): and, in general, if we have found \(x + a = b\), where \(a\) and \(b\) express any known numbers, we have only to subtract \(a\) from both sides, to obtain the equation \(x=b-a\), which indicates the value of \(x\).

575 If we have the equation \(x-a=b\), we must add \(a\) to both sides, and shall obtain the value of \(x=b +a\). We must proceed in the same manner, if the equation have this form, \(x-a=a^2+1\): for we shall immediately find \(x=a^2 +a+ 1\).

In the equation \(x-8a=20-6a\), we find \(x=20-6a+8a\), or \(x=20+2a\).

And in this, \(x+6a=20+3a\), we have \(x=20+3a-6a\), or \(x=20-3a\).

576 If the original equation have this form, \(x - a +b=c\), we may begin by adding \(a\) to both sides, which will give \(x + b =c + a\); and then subtracting \(b\) from both sides, we shall find \(x = c +a - b\). But we might also add \(+ a - b\) at once to both sides; and thus obtain immediately \(x=c+a-b\).

So likewise in the following examples:

If \(x-2a+3b=0\), we have \(x=2a-3b\).

If \(x-3a+2b=25+a+2b\), we have \(x=25+4a\).

If \(x-9 +6a=25+2a\), we have \(x=34-4a\).

577 When the given equation has the form \(ax=b\), we only divide the two sides by \(a\), to obtain \(x = \dfrac{b}{a}\). But if the equation have the form \(ax+b-c=d\), we must first make the terms that accompany \(ax\) vanish, by adding to both sides \(- b + c\); and then, dividing the new equation \(ax=d-b+c\) by \(a\), we shall have \(x = \dfrac{d-b+c}{a}\).

The same value of \(x\) would have been found by subtracting \(+ b -c\) from the given equation: that is, we should have had, in the same form,

\[ax = d - b + c, \quad \textrm{and} \quad x = \dfrac{d-b+c}{a}.\]Hence,

If \(2x+5=17\), we have \(2x=12\), and \(x=6\).

If \(3x-8=7\), we have \(3x= 15\), and \(x=5\).

If \(4x - 5 - 3a = 15 + 9a\), we have \(4x = 20 + 12a\),
and consequently \(x=5 + 3a\).

578 When the first equation has the form \(\dfrac{x}{a}=b\), we multiply both sides by \(a\), in order to have \(x = ab\).

But if it is \(\dfrac{x}{a} + b - c = d\), we must first make \(\dfrac{x}{a} = d- b + c\); after which we find

\[x=(d -b + c)a=ad-ab + ac.\]Let \(\frac{1}{2}x-3=4\), then \(\frac{1}{2}x=7\), and \(x= 14\).

Let \(\frac{1}{3}x - 1 + 2a = 3 + a\), then \(\frac{1}{3}x = 4 - a\), and
\(x=12-3a\).

Let \(\dfrac{x}{a-1}-1=a\), then \(\dfrac{x}{a-1} =a + 1\), and \(x = a^2 - 1\).

579 When we have arrived· at such an equation as \(\dfrac{ax}{b} = c\), we first multiply by \(b\), in order to have \(ax =bc\), and then dividing by \(a\), we find \(x = \dfrac{bc}{a}\).

If \(\dfrac{ax}{b}- c =d\), we begin by giving the equation this form, \(\dfrac{ax}{b} = d + c\); after which, we obtain the value of \(ax = bd + bc\), and then that of \(x = \dfrac{bd+bc}{a}\).

Let \(\frac{2}{3}x - 4 = 1\), then \(\frac{2}{3}x = 5\), and \(2x = 15\); whence \(x=\frac{15}{2}=7\frac{1}{2}\).

If \(\frac{3}{4}x+\frac{1}{2}=5\), we have \(\frac{3}{4}x =5 -\frac{1}{2} =\frac{9}{2}\); whence \(3x=18\), and \(x=6\).

580 Let us now consider a case, which may frequently occur; that is, when two or more terms contain the letter \(x\), either on one side of the equation, or on both.

If those terms are all on the same side, as in the equation \(x+\frac{1}{2}x+5=11\), we have \(x+\frac{1}{2}x=6\); or \(3x=12\); and lastly, \(x=4\).

Let \(x + \frac{1}{2}x + \frac{1}{3}x = 44\), be an equation, in which the value of \(x\) is required. If we first multiply by 3, we have \(4x + \frac{3}{2}x = 132\); then multiplying by 2, we have \(11x =264\); wherefore \(x=24\). We might have proceeded in a more concise manner, by beginning with the reduction of the three terms which contain \(x\) to the single term \(\frac{11}{6}x\); and then dividing the equation \(\frac{11}{6}x=44\) by 11. This would have given \(\frac{1}{6}x=4\), and \(x=24\), as before.

Let \(\frac{2}{3}x - \frac{3}{4}x + \frac{1}{2}x =1\). We shall have, by reduction, \(\frac{5}{12}x=1\), or \(5x=12\), and \(x=2\frac{2}{5}\).

And, generally, let \(ax - bx + cx = d\); which is the same as \((a-b+c)x=d\), and, by division, we derive \(x=\dfrac{d}{a-b+c}\).

581 When there are terms containing \(x\) on both sides of the equation, we begin by making such terms vanish from that side from which it is most easily expunged; that is to say, in which there are the fewest terms so involved.

If we have, for example, the equation \(3x+2=x+ 10\), we must first subtract \(x\) from both sides, which gives \(2x+2= 10\); wherefore \(2x=8\), and \(x=4\).

Let \(x + 4 = 20 - x\); here it is evident that \(2x + 4 = 20\); and consequently \(2x= 16\), and \(x=8\).

Let \(x+8=32-3x\), this gives us \(4x+8=32\); or \(4x=24\), whence \(x=6\).

Let \(15-x=20-2x\), here we shall have \(15 +x=20\), and \(x=5\).

Let \(1 +x=5-\frac{1}{2}x\); this becomes \(1 +\frac{3}{2}x=5\), or \(\frac{3}{2}x=4\); therefore \(3x=8\); and lastly, \(x=\dfrac{8}{3}=2\frac{2}{3}\).

If \(\frac{1}{2}-\frac{1}{3}x=\frac{1}{3}-\frac{1}{4}x\), we must add \(\frac{1}{3}x\), which gives \(\frac{1}{2}=\frac{1}{3} +\frac{1}{12}x\); subtracting \(\frac{1}{3}\), and transposing the terms, there remains \(\frac{1}{12}x=\frac{1}{6}\); then multiplying by 12, we obtain \(x=2\).

If \(1\frac{1}{2}-\frac{2}{3}x=\frac{1}{4}+\frac{1}{2}x\), we add \(\frac{2}{3}x\), which gives \(1\frac{1}{2}=\frac{1}{4}+\frac{7}{6}x\); then subtracting \(\frac{1}{4}\), and transposing, we have \(\frac{7}{6}x= 1\frac{1}{4}\), whence, by multiplying by 6 and dividing by 7, we deduce \(x=1\frac{1}{14}=\frac{15}{14}\).

582 If we have an equation in which the unknown number \(x\) is a denominator, we must make the fraction vanish by multiplying the whole equation by that denominator.

Suppose that we have found \(\frac{100}{x} -8= 12\), then, adding 8, we have \(\frac{100}{x} = 20\); and multiplying by \(x\), it becomes \(100=20x\); lastly, dividing by 20, we find \(x=5\).

Let now \(\dfrac{5x+3}{x-1} = 7\); here, multiplying by \(x -1\), we \(5x+3=7x-7\); and subtracting \(5x\), there remains \(3 =2x-7\); then adding 7, we have \(2x= 10\); whence \(x=5\).

583 Sometimes, also, radical signs are found in equations of the first degree. For example: A number \(x\), below 100, is required, such, that the square root of \(100-x\) may be equal to 8; or \(\surd(100-x)=8\). The square of both sides will give \(100-x=64\), and adding \(x\), we have \(100=64+x\): whence we obtain \(x=100-64=36\).

Or, since \(100-x=64\), we might have subtracted 100 from both sides: which would have given \(-x= -36\); or, multiplying by \(-1\), \(x=36\).

584 Lastly, the unknown number \(x\) is sometimes found as an exponent, of which we have already seen some examples; and, in this case, we must have recourse to logarithms.

Thus, when we have \(2^x=512\), we take the logarithms of both sides; whence we obtain \(x \log 2 = \log 512\); and dividing by \(\log 2\), we find \(x = \dfrac{\log 512}{\log 2}\). The Tables then give,

\[x = \dfrac{2.7092700}{0.3010300} = \dfrac{27092700}{3010300},\]or \(x=9\).

Let \(5 \cdot 3^{2x} - 100 = 305\); we add 100, which gives \(5 \cdot 3^{2x} = 405\); dividing by 5, we have \(3^{2x}= 81\); and taking the logarithms, \(2x \log 3=\log 81\), and dividing by \(2 \log 3\), we have

\[x = \dfrac{\log 81}{2 \log 3},\]or,

\[x = \dfrac{\log 81}{\log 9};\]whence

\[x = \dfrac{1.9084850}{0.9542425}=\dfrac{19084850}{09542425}=2.\]#### Editions

- Leonhard Euler.
*Elements of Algebra*. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822. - Leonhard Euler.
*Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange.*Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.