Simplify:

15π‘₯squared𝑦 Γ· π‘₯𝑦 + 3π‘₯ Γ— βˆ’2𝑦 βˆ’ 10 Γ— π‘₯ Γ— 2

asked by guest
on Nov 23, 2024 at 1:20 am



You asked:

Simplify the expression: 15x2yxy+3xβ‹…βˆ’(2y)βˆ’10xβ‹…2\frac{15 {x}^{2} y}{x y} + 3 x \cdot -\left( 2 y \right) - 10 x \cdot 2

MathBot Answer:

Simplified:

15x2yxy+3xβ‹…βˆ’(2y)βˆ’10xβ‹…2=x(βˆ’6yβˆ’5) \frac{15 {x}^{2} y}{x y} + 3 x \cdot -\left( 2 y \right) - 10 x \cdot 2 = x \left(- 6 y - 5\right)

15yx2xy+(βˆ’1)3β‹…2yxβˆ’2β‹…10x=15yx2xy+(βˆ’1)3β‹…2yxβˆ’20x=15yx2xyβˆ’20xβˆ’6xy=βˆ’20x+15xβˆ’6xy=βˆ’5xβˆ’6xy=x(βˆ’6yβˆ’5)\begin{aligned}\frac{15 y x^{2}}{x y} + \left(-1\right) 3 \cdot 2 y x - 2 \cdot 10 x&=\frac{15 y x^{2}}{x y} + \left(-1\right) 3 \cdot 2 y x - 20 x\\&=\frac{15 y x^{2}}{x y} - 20 x - 6 x y\\&=- 20 x + 15 x - 6 x y\\&=- 5 x - 6 x y\\&=x \left(- 6 y - 5\right)\end{aligned}

Factored:

15x2yxy+3xβ‹…βˆ’(2y)βˆ’10xβ‹…2=βˆ’x(6y+5) \frac{15 {x}^{2} y}{x y} + 3 x \cdot -\left( 2 y \right) - 10 x \cdot 2 = - x \left(6 y + 5\right)


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