Question

. Subtract 5x2 – 6y2 + 8y – 5 from 7x2 – 5xy + 10y2 + 5x – 4y

Answer 1

Step-by-step explanation:

Given:-

Given expressions are :

5x^2 – 6y^2 + 8y – 5 and

7x^2 – 5xy + 10y^2 + 5x – 4y

To find:-

Subtract 5x^2 – 6y^2 + 8y – 5 from

7x^2 – 5xy + 10y^2 + 5x – 4y

Solution:-

Method -1:-

On Subtracting 5x^2 – 6y^2 + 8y – 5 from

7x^2 – 5xy + 10y^2 + 5x – 4y

=>(7x^2 – 5xy + 10y^2 + 5x – 4y)-(5x^2 – 6y^2 + 8y – 5 )

=>7x^2 – 5xy + 10y^2 + 5x – 4y- 5x^2 + 6y^2 - 8y +5

=>(7x^2-5x^2)+(10y^2+6y^2)+(-5xy)+(5x)+(-8y-4y)+5

=>2x^2 + 16y^2 -5xy +5x -12y +5

Method-2:-

7x^2 -5xy +10y^2 +5x -4y

5x^2+0xy - 6y^2 +0x +8y -5

(-)

________________________

2x^2 -5xy +16y^2 +5x -12y +5

_________________________

Answer:-

The answer for the given problem is

2x^2 +1 6y^2 - 5xy + 5x -12y + 5

Answer 2

${\large{\pmb{\sf{Correct \; question...}}}}$

$\sf Subtract = \: 5x^{2} - 6y^{2} + 8y - 5 \: from \: 7x^{2} - 5xy + 10y^{2} + 5x - 4y$

${\large{\pmb{\sf{Full \; Solution...}}}}$

~ As it's given that we have to ${\sf{Subtract = \: 5x^{2} - 6y^{2} + 8y - 5 \: from \: 7x^{2} - 5xy + 10y^{2} + 5x - 4y}}$ Now let's carry on.

$:\implies \sf 5x^{2} - 6y^{2} + 8y - 5 \\ \\ :\implies \sf 7x^{2} - 5xy + 10y^{2} + 5x - 4y \\ \\ \bf Now \: let's \: write \: them \: in \: the \: form \: of \: like \: terms \\ \\ :\implies \sf 5x^{2} - 6y^{2} + 8y - 5 \dots Equation \: 1 \\ \\ :\implies \sf 7x^{2} + 10y^{2} - 4y -5xy + 5x \dots Equation \: 2 \\ \\ \bf Now \: let's \: subtract \: given \: numbers \\ \\ \bf We \: have \: to \: subtract \: Equation \: 2 \: from \: Equation \: 1 \\ \\ :\implies \sf 7x^{2} + 10y^{2} - 4y -5xy + 5x - (5x^{2} - 6y^{2} + 8y - 5) \\ \\ :\implies \sf 7x^{2} + 10y^{2} - 4y -5xy + 5x - 5x^{2} + 6y^{2} + 8y - 5 \\ \\ :\implies \sf 2x^{2} + 16y^{2} + 4y - 5xy + 5x - 5$

${\large{\pmb{\sf{Additional \; information...}}}}$

❶ Linear equation = Here we will have to deal with linear expressions in just one variable. Such equations are known to be “linear equation in one variable”

❷ An algebraic equation in an equality involving variable. It has an equality sign. The expression on the left of equality sign is LHS. The expression on the right of equality sign is RHS like in expression 2x - 3 = 7

↦ 2x is variable

↦ = is the sign of equality

↦ 7 is equation

↦ 2x - 3 is LHS

↦ 7 is RHS..

❸ In an equation the values, of the expression on LHS and RHS are equal. This happen to be true ! for certain values of that variable. The values are the solution of that equation like,

↦ x = 5 is the solution of the equation

↦ 2x - 3 = 7 [ x = 5 ]

↦ LHS = 2 × 5 - 3 = 7 = RHS

↦ On the other hand, x = 10 is nor a solution of the equation [ x = 10 ]

↦ LHS = 2 × 10 - 3 = 17

↦ This isn't equal to the RHS.

❹ How to find solution for equation?

We have to assume that the 2 sides of the equation are in a balance. We have to perform the same mathematical operation on both sides of the equation so that the balance isn't disturbed. A few such steps give you your solution always....!!