Order operations and algebra - class-VII

To subtract the first expression from the second, compute ((1/3)y^3 + (5/7)y^2 - 2) - (-(2/3)y^3 - (2/7)y^2 - 5). Distributing the negative sign gives ((1/3) + (2/3))y^3 + ((5/7) + (2/7))y^2 + (-2 + 5), which simplifies to y^3 + y^2 + 3.

$Multiplying\quad (3y+4z)(3y-4z),\quad (2y+7z)(y+z),\ =-16z^{ 2 }+12yz-12yz+9y^{ 2 }\quad +\quad 7z^{ 2 }+7yz+2yz+2y^{ 2 }\ =-9z^{ 2 }+9yz+11y^{ 2 }$

$(a+b)(c-d)+(a-b)(c+d)+2(ac+bd)$
$=a(c-d)+b(c-d)+a(c+d)-b(c+d)+2(ac+bd)$
$=ac-ad+bc-bd+ac+ad-bc-db+2ac+2bd$
$=4ac$

$(x + y)(x^2 - xy + y^2)$
$=x(x^2 - xy + y^2)+y(x^2 - xy + y^2)$
$=x^3-x^2y+xy^2+x^2y^-xy^2+y^3$
$=x^3+y^3$

The given expression $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$ can be simplified as follows:

Given 4x + y = 10, if x increases by 3, the new equation is 4(x + 3) + y_new = 10. This expands to 4x + 12 + y_new = 10, which simplifies to y_new = 10 - 4x - 12. Since the original y = 10 - 4x, the new y is y - 12.

$\sqrt {13+\sqrt {44+10^2}}\$

$x^2+y^2=100\y=6\x^2+6^2=100\x^2=100-36\x^2=64\x=\pm 8$

Given,
$ b = 6 - \left[ \frac { 4b+3 }{ 2a-5 }  \right]  $
$ => \left[ \frac { 4b+3 }{ 2a-5 }  \right] = 6 - b $
$ => 4b + 3 = (6-b)(2a-5) $
$ => 4b + 3 = 12a - 30 - 2ab +5b $
$ => 12a - 33  -2ab + b = 0 $
$ =>  a(12 -2b) = 33 - b $
$ => 2a(6-b) = 33-b $
$ => a = \frac {33-b}{2(6-b)} $

Given, $ c = \frac {3b-4}{4b + 3} $

But, $ b = \frac {2a}{a-2} $
Hence, $ c = \frac {3(\frac {2a}{a-2} )-4}{4(\frac {2a}{a-2} ) + 3} $
$ => c = \frac {6a-4a + 8}{8a +3a - 6} $
$ => c = \frac {2a+8}{11a -6} $

We apply BODMAS rule to find the value of the given expression. According to BODMAS rule, if an expression contains brackets we have to first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left to right.

Given exp.$\displaystyle \frac{2}{1+\frac{1}{\frac{1}{2}}}\times\frac{3}{\frac{15}{12}\div \frac{5}{4}}=\frac{2}{1+2}\times\frac{3}{\frac{15}{12}\times\frac{4}{5}}=\frac{2}{3}\times\frac{3}{1}=2$

$(a-b)^2=a^2+b^2-2ab$

$((x^3-2)\div2^2)\times 4+16$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $((x^3-2)\div2^2)\times 4+16$
$=$ $\dfrac{x^3-2}{4}\times 4+16$
$=$ $x^3+14$

$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
$=$ $3x.x^2+3x.1-[2x.x^2+2x.x-2x.1+1]-x^2$
$=$ $3x^3+3x-2x^3-2x^2+2x-1-x^2$
$=$ $x^3-3x^2+5x-1$

$4-x^2\div x +(4\times-(\dfrac{2x^3}{x^2}))-3^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $4-x^2\div x +(4\times-(\dfrac{2x^3}{x^2}))-3^2$
$=$ $4-x+(4\times -2x)-9$
$=$ $4-x-8x-9$
$=$ $-9x-5$

$[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
$=$ $\dfrac{(100+x)(x^4)}{x^2}\times 2-x-x^2+1$
$=$ $200x^2+2x^3-x-x^2+1$
$=$ $2x^3+199x^2-x+1$

$x-1[(x^2+x-2)(x^2-1^2)\div (x-1)^2]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $\frac{x-1(x^2+x-2)(x+1)(x-1)}{(x-1(x-1)}$
$=$ $x^3+x^2-2x+x^2+x-2$
$=$ $x^3+2x^2-x-2$

$12-[5y+2x(y^2-2x+2)+6y-(y^2-1)]\times 0.1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $12-[5y+2xy^2-4x^2+4x+6y-y^2+1]\times 2$
$=$ $12- [2xy^2-4x^2+4x+11y-y^2+1]\times 2$
$=$ $12-[4xy^2-8x^2+8x+22y-2y^2+2]$
$=$ $12-4xy^2+8x^2-8x-22y+2y^2-2$
$=$ $8x^2+2y^2-4xy^2-8x-22y+10$

$x^2-x[(-x)(-2+x)]\div x+x^3-3x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $-2x^2-x[\dfrac{(-x)(-2+x)}{x}]+x^3$
$=$ $-2x^2-x[2-x]+x^3$
$=$ $x^3-2x^2-2x+x^2$
$=$ $x^3-x^2-2x$

$2y-1(y-y^2)+5y[(-2y)(y^2-1)]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $2y-y+y^2+5y[-2y^3+2y]$
$=$ $y+y^2-10y^4+10y^2$
$=$ $-10y^4+11y^2+y$

$4x^3[(3x-x^2)-1]+(x^2)[x+1]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $4x^3[(3x-x^2)-1]+(x^2)[x+1]$
$=$ $4x^3[3x-x^2-1]+x^3+x^2$
$=$ $12x^4-4x^5-4x^3+x^3+x^2$
$=$ $-4x^5+12x^4-3x^3+x^2$

$5x[2x(x^2+x^3)-x^3]-4x^2\div x^2-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $5x[2x(x^2+x^3)-x^3]-4x^2\div x^2-12x$
$=$ $5x[2x^3+2x^4-x^3]-\dfrac{4x^2}{x^2}-12x$
$=$ $5x[x^3+2x^4]-4-12x$
$=$ $5x^4+10x^4-12x-4$
$=$ $15x^4-12x-4$

$xy^2+x^3-x[x^2-x][2x]+(x-1)x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $xy^2+x^3+[-x^3+x^2]2x+x^3-x^2$
$=$ $xy^2+x^3-x^4+2x^3+x^3-x^2$
$=$ $xy^2+4x^3-x^4-x^2$
$=$ $-x^4+4x^3-x^2+xy^2$

$x^2\times(x-1)+[(2x+2)\times 4x]-1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $x^2\times(x-1)+[(2x+2)\times 4x]-1$
$=$ $x^3-x^2+8x^2+8x-1$
$=$ $x^3+7x^2+8x-1$

$24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
$=$ $24x+24-x^2+24-x-2$
$=$ $-x^2+46+23x$
$=$ $-x^2+23x+46$

$3x(x-2)+x(x^2\times 2x)-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $3x(x-2)+x(x^2\times 2x)-12x$
$=$ $3x^2-6x+x^3\times2x^2-12x$
$=$ $3x^2-6x+2x^5-12x$
$=$ $2x^5+3x^2-18x$

$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$

Given $ \frac {100 - m}{100} \times x = \frac {100 + m}{100} \times y $
$ => 100x - mx = 100y + my $
$ => 100x - 100y = mx + my $
$ => 100 (x-y) = m (x + y) $
$ => m = \frac {100(x-y)}{x+y} $

When , $ 2x = 3y => x = \frac {3y}{2} $
We have, $ m = \frac {100(\frac {3y}{2} -y)}{\frac {3y}{2} +y} $
$ => m =\frac {100(\frac {y}{2})}{\frac {5y}{2}} $
$ => m =\frac {100}{5} = 20 $

Given, $ a = \frac {1 +{b}^{2}}{1 - {b}^{2}} $
$ => a -a{b}^{2} = 1 +{b}^{2} $
$ => a- 1 = {b}^{2} (1 + a) $
$ =>{b}^{2} = \frac {a-1}{1+a} $