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Ipatiy [6.2K]
3 years ago
10

4x+1=9+4y please help me solve for x please show step by step I’m very confused

Mathematics
2 answers:
Inessa05 [86]3 years ago
7 0

Answer:

<em>x - y = 2</em>

Step-by-step explanation:

First of all, you would combine like terms, giving you 4x - 4y + 1 = 9.

Next, subtract the numbers, which gives you 4x - 4y = 8.

Finally, divide both sides by 4, which will come to the conclusion of x - y = 2, the simplest form of which this equation can be broken down to.

ale4655 [162]3 years ago
4 0

Answer:

x =  y + 2

Step-by-step explanation:

given

4x + 1 = 9 + 4y

4x = 4y + 9 - 1

4x = 4y + 8

making x the subject by dividing through by 4

\frac{4x}{4}  =  \frac{4y + 8}{4}

x =  \frac{4y + 8}{4}

x =  \frac{4y}{4}  +  \frac{8}{4}

x = y + 2

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Answer:

1 \frac{1}{2}

Step-by-step explanation:

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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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3 years ago
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