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Svetlanka [38]
3 years ago
7

Which of these strategies would eliminate a variable in the system of equations? \begin{cases} 6x + 5y = 1 \\\\ 6x - 5y = 7 \end

{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ 6x+5y=1 6x−5y=7 ​
Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

x = 0.667 and y = -0.6

Step-by-step explanation:

Given that,

6x + 5y = 1 ....(1)

6x - 5y = 7 ...(2)

We need to solve the above equations.

Subtract equations (1) and (2)

6x + 5y -(6x - 5y) = 1-7

6x+5y-6x+5y = -6

10y=-6

y = -0.6

Put the value of y in equation (1).

6x + 5(-0.6) = 1

6x = 1-5(-0.6)

6x = 4

x = 0.667

Hence, first step is to subtract equation (1) and (2) then put the value of y in equation (1).

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Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2
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\bold{[ \ Answer \ ]}

\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}

\bold{[ \ Explanation \ ]}

  • \bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}

\bold{-------------------}

  • \bold{Rewrite}

\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}

  • \bold{Rewrite}

\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}

  • \bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}

\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}

  • \bold{Rewrite}

\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}

  • \bold{Simplify}

\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}

\boxed{\bold{[] \ Eclipsed \ []}}

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3 years ago
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Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).
Serjik [45]

Answer:

\displaystyle f(x)=x^2+2x+2

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

\displaystyle f(x)=ax^2+bx+c

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

\displaystyle f(-1)=a(-1)^2+b(-1)+c

\displaystyle f(-1)=a-b+c

\displaystyle a-b+c=1.....[eq\ 1]

Now we use the second condition f(1)=5

\displaystyle f(1)=a(1)^2+b(1)+c

\displaystyle f(1)=a+b+c

\displaystyle a+b+c=5.......[eq\ 2]

Finally, we use the third condition f(2)=10

\displaystyle f(2)=a(2)^2+b(2)+c

\displaystyle f(2)=4a+2b+c

\displaystyle 4a+2b+c=10....[eq\ 3]

We put together eq 1, eq 2, and eq 3 to form the system

\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.

Adding the first two equations we have

\displaystyle 2a+2c=6

\displaystyle a+c=3

And also

\displaystyle b=2

Using the above equation and the value of b in the third equation, we have

\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.

Subtracting the first equation from the second

\displaystyle 3a=3

\displaystyle a=1

And therefore

\displaystyle c=2

Now we have all the values, the quadratic function is

\displaystyle \boxed{f(x)=x^2+2x+2}

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