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romanna [79]
3 years ago
12

The sum of four times a number and five times another number is -28. The sum of nine times the first number and two times the se

cond number is 11
Mathematics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

<h2>3 and -8</h2>

Step-by-step explanation:

x,\ y-the\ numbers\\\\\text{The sum of four times a number and five times another number is -28:}\\\\4x+5y=-28\\\\\text{The sum of nine times the first number and two times the second number is 11:}\\\\9x+2y=11\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}4x+5y=-28&\text{multiply both sides by2}\\9x+2y=11&\text{multiply both sides by (-5)}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}8x+10y=-56\\-45x-10y=-55\end{array}\right}\qquad\text{add both sides of the equation}

.\qquad-37x=-111\qquad\text{divide both sides by (-37)}\\.\qquad\qquad\boxed{x=3}\\\\\text{Put the value of x to the first equation:}\\\\4(3)+5y=-28\\12+5y=-28\qquad\text{subtrac 12 from both sides}\\5y=-40\qquad\text{divide both sides by 5}\\\boxed{y=-8}

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\implies {\blue {\boxed {\boxed {\purple {\sf {   \: 28 {x}^{3}  + 48 {x}^{2}  + 62x + 30}}}}}}

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}

(4 {x}^{2}  +  4x + 6)(7x + 5)

= \: 7x(4 {x}^{2}  + 4x + 6) + 5(4 {x}^{2}  + 4x + 6)

=  \: 28 {x}^{3}  + 28 {x}^{2}  + 42x + 20 {x}^{2}  + 20x + 30

Combining like terms, we have

=  \: 28 {x}^{3}  + (28 {x}^{2}  + 20 {x}^{2} ) + (42x + 20x) + 30

=  \: 28 {x}^{3}  + 48 {x}^{2}  + 62x + 30

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

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Step-by-step explanation:

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