Question

and y=40 (2x+3y) (2x+1)
Multiply and verify for x = 1​

Answer 1

Given expression:

(2x+3y) (2x+1)

We are also given that, x = 1 and y = 40.

Solution:

First we will simplify this expression: (2x+3y) (2x+1)

2x(2x+1) + 3y(2x+1) [Using distributive property]

$\boxed {\sf {\blue{4x² + 2x + 6xy + 3y}}}$ ---> This is the simplified expression.

Now we will put the values of x and y in the expression obtained.

x = 1

y = 40

4x² + 2x + 6xy + 3y

4×(1)² + 2×1 + 6×1×40 + 3×40

4×1+2+6×40+120

4+2+240+120

366

Verification:

On substituting the value of x and y as 1 and 40 respectively in the given expression:

(2x+3y) (2x+1)

(2×1+3×40) (2×1+1)

(2+120) (2+1)

122×3

= 366

366 = 366

LHS = RHS

Hence Verified!

Final answer:

Therefore the result of the expression (2x+3y) (2x+1) is $\bf \red{366}$ (given that x = 1 and y = 40)

Answer 2

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Given:

• Equation: $\sf (2x+3y)(2x+1)$
• $\sf x=1$
• $\sf y=40$

To do:

• Simplify the equation
• Find its solution
• Verify

Solution:

$\normalsize{\underline{\bold{1)\: Simplifying\: the\:given\: equation:}}}$

$\sf\implies (2x + 3y)(2x + 1)$

$\sf\implies 2x(2x + 1) + 3y(2x + 1)$

• $\large{\underline{\boxed{\sf{\red{{4x}^{2} + 2x + 6xy + 3y}}}}}$

$\rightarrow$For simplifying, we have used "Distributive Property".

$\normalsize{\underline{\bold{2)\: Finding\:its\:solution:}}}$

$\rightarrow$Insert the given values of in place of x and y in the simplified equation.

$\sf\implies{{4x}^{2} + 2x + 6xy + 3y}$

$\sf\implies{4×{1}^{2} + 2×1 + 6×1×40 + 3×40}$

$\sf\implies 4×1+2+6×40+3×40$

$\sf\implies 4+2+240+120$

$\sf\implies 6+360$

• $\large\underline{\boxed{\sf{\red{366}}}}$

$\normalsize{\underline{\bold{3)\: Verification:}}}$

To verify, substitute the given values of 'x' and 'y' respectively in the given equation.

$\sf\implies (2x+3y)(2x+1)$

$\sf\implies (2×1+3×40)(2×1+1)$

$\sf\implies (2+120)(2+1)$

$\sf\implies 122×3$

$\sf\implies{\red{366}}$

$\bold{\implies L.H.S = R.H.S}$

• $\large{\underline{\underline{\tt{\green{Hence,\: verified\:!}}}}}$

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