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3 years ago

I need help with elimination with substitution. The equations are 3x+4y=19 and 3x+6y=33

1 answer:

4 0

For elimination, multiply one whole equation by negative one (-1), then add or subtract according to your signs. After that, it will be a one-step equation.

3x + 4y = 19 3x + 4y = 19 3x + 4y = 19 -2y = -14 y = 7
3x + 6y = 33 -1 (3x + 6y = 33) -3x - 6y = -33 -14 / -2

Then you would go back and substitute the value of (y) back into either equation and then solve for the remaining variable (x). Finally, use both values to make an ordered pair.

3x + 4y = 19 3x = -9 x = -3 (-3 , 7)
3x + 4(7) = 19 (-9 / 3)
3x + 28 = 19

Good Luck

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Tju [1.3M]

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$\displaystyle \frac{x^4 + 10x^3 + 25x^2 + 3x - 24}{x^2 + 5x - 4} = x^2 + 5x - 4 + \frac{3x - 8}{x^2 + 5x - 4}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding

Algebra II

Polynomial Division

Long Division
Synthetic Division

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{x^4 + 10x^3 + 25x^2 + 3x - 24}{x^2 + 5x - 4}$

Step 2: Long Division

See attachment.

Multiply quotient a and divisor, then subtract from dividend: $\displaystyle x^2(x^2 + 5x - 4) = x^4 + 5x^3 - 4x^2 \leftarrow \text{red 1}$
Multiply quotient b and divisor, then subtract from new dividend: $\displaystyle 5x(x^2 + 5x - 4) = 5x^3 + 25x^2 - 20x \leftarrow \text{red 2}$
Multiply quotient c and divisor, then subtract from new dividend: $\displaystyle 4(x^2 + 5x - 4) = 4x^2 + 20x - 16 \leftarrow \text{red 3}$
Write remainder: $\displaystyle \frac{r(x)}{b(x)} = \frac{3x - 8}{x^2 + 5x - 4}$

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