simplify each of the
 following algebraic expressions by removing grouping symbols.
 ` x+[5y-{2x-(3y-5x)}]`

 `2a-[4b-{4a-3(2a-b)}]`
step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1. hint: mathematics includes the study of topics which are related to quantity, structure, space and change. to solve this problem, we first try to simplify the expression. for doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. in this way we can easily factorise the expression.complete step-by-step answer:according to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. to simplify this complex expression, we let ${{x}^{2}}-4x=t$.this can be mathematically expressed as: $t(t-1)-20$.since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:${{t}^{2}}-t-20$now, by using the middle term splitting method of separating the variables, we get$\begin{align}  & {{t}^{2}}+4t-5t-20=0 \\  & t(t+4)-5(t+4)=0 \\  & (t-5)(t+4)=0 \\ \end{align}$ so, the factors of the expression in terms of t are (t-5) and (t+4).now, finally replacing t as ${{x}^{2}}-4x$,we get$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$ on further simplifying this expression, by using middle term splitting method again in part one:$\begin{align}  & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\  & \rightarrow x(x+1)-5(x+1)=0 \\  & (x-5)(x+1)=0 \\ \end{align}$ now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get${{x}^{2}}-4x+4={{(x-2)}^{2}}$$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$ hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$ note: this question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. but the complexity of the problem would increase by doing so. hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer. 2x²+3x+5=0 has solutions by the quadratic formula x=(-3±√(9-4×2×5))/4 =(-3±√(-31))/4
there are no real solutions, since the discriminant is -31, whose square root is imaginary. slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4