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Any equation may be cleared of fractions, by multiplying each of its terms, successively, by the denominators of those fractions, or by multiplying both sides by the product of all the denominators, or by any quantity that is a multiple of them.
Thus, if+=5, then, multiplying by 3, we have +
3x 15; and this, multiplied by 4, gives 4x+3x=60;
whence, by addition, 7x=60, or x= =8
And, if += 10; then, multiplying by 12, (which
is a multiple of 4 and 6,) 3x+2x=120, or 5x=120, or
It also appears, from this rule, that if the same number, or quantity, be found in each of the terms of an equation, either as a multiplier or divisor, it may be expunged from all of them, without altering the result.
Thus, if ax-ab+ac; then, by cancelling, x=b+c.
And, if +
-=- ; then x+b=c, or x=c-b.
If the unknown quantity, in any equation, be in the form of a surd, transpose the terms so that this may stand alone, on one side of the equation, and the remaining terms on the other (by Case 1); then involve each of the sides to such a power as corresponds with the index of
the surd, and the equation will be rendered free from any irrational expression.
Thus, if x-2=3; then will √x=3+2=5, or, by squaring, x52=25.
And if 3x+4=5; then will 3x+4=25, or 3x=
Also, if /2x+3+4=8; then will 3/2x+3=8—4 =4, or 2x+3=43=64, and consequently 2x=64-3 =61, or x=
If that side of the equation, which contains the unknown quantity, be a complete power, the equation may be reduced to a lower dimension, by extracting the root of the said power on both sides of the equation.
Thus, if x2=81; then x=81=9; and if x3=27, then x= =3/27=3.
Also, if 3x2-9=24; then 3x2=24+9=33, or x2= 11, and consequently x=11.
And, if x+6x+9=27; then, since the left hand side of the equation is a complete square, we shall have, by extracting the roots, x+3=27=✅✔✅9×3=3,✓/3, or x=3√3-3.
Any analogy, or proportion, may be converted into an equation, by making the product of the two extreme terms equal to that of the two means.
Thus, if 3x 16: 5: 6; then 3x X6=16X5, or
Also, if 12 -x: :: 41; then 12-x=.
or 2x+x=12; and consequently x=
1. Given 5x-15=2x+6, to find the value of x. Here 5x-2x-6+15, or 3x=6+15=21; and there
2. Given 40-6x-16-120-14x, to find the value of x.
Here 14x-6x=120–40+16; or 8x = 136–40 = 96; and therefore x=12,
3. Given 3x2-10x=8x+x2, to find the value of x. Here 3x-10=8+x, by dividing by x; or 3x-x=8 +10=18, by transposition.
And consequently 2x=18, or x===
4. Given 6ax3-12abx2-3ax3+6ax2, to find the value of x.
Here 2x-46x+2, by dividing by 3ax2; or 2x-x 246; and therefore x=4b+2.
5. Given x2+2x+1=16, to find the value of x. Here x+14, by extracting the square root of each side.
And therefore, by transposition, x=4-13.
6. Given 5ar-3b-2dx+c, to find the value of x.
Here 5ax-2dx=c+3b; or (5α-2d)x=c+36; and
therefore, by division, x=.
12x-8x+6x=240; whence 10x=240, or x=24.
Here x-3+40-x+19; or 3x-9+2x=120—
3x+57; whence 3x+2x+3x=120 + 57+9; that is 8x=186, or x=231.
9. Given +5=7, to find the value of x.
Here✓7-5=2; whence, by squaring,
=4, and 2x=12, or x=6.
10. Given x + √(a2+x3) :
value of x.
to find the
Here x/(a+x2)+a2+x2 = 2a2; or x/(a2+x2)= a2 - x2,and x2 (a2+x2)=a1 — 2a2x2+x4; whence a2x2 + x1=aa — 2α2x2+x1‚anda2x2=a4-2a2x2; therefore3a2x2 a4 a2
=α, or x2=