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stealth61 [152]
2 years ago
14

Which shows the correct first step to solving the system of equations in the most efficient manner? 3 x 2 y = 17. X 4 y = 19. X

= negative 4 y 19 x = StartFraction negative 2 y 17 Over 3 EndFraction 4 y = negative x 19 y = StartFraction negative 3 x 17 Over 2 EndFraction.
Mathematics
2 answers:
Shkiper50 [21]2 years ago
4 0

The correct first step to solving the system of equations in the most efficient manner is x = -4y +19.

The given system of equations is;

\rm 3 x + 2 y = 17\\\\x + 4 y = 19

<h2>What is the first step to solving an equation?</h2>

The first step would be to rearrange the variables in the second equation so we have a defined expression for x.

We would want to do this so we could then plug in the defined expression for x into the other equation and solve for the other variable.

x + 4y = 19

Rearrange this by getting the 4y to the other side.

We can get 4y on the other side by subtracting 4y from both sides

x + 4y - 4y = 19 - 4y

x = -4y + 19

Hence, the correct first step to solving the system of equations in the most efficient manner is x = -4y +19.

To know more about the System of equations click the link given below.

brainly.com/question/15240525

WINSTONCH [101]2 years ago
4 0

Answer:

(A)x = negative 4 y + 19

Step-by-step explanation:

i did the test on edge good luck

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3 years ago
One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

Cross Multiply

9dp = c^2

or

c^2 = 9dp

Hence, the relationship between d, c and p is: c^2 = 9dp

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Answer:

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Step-by-step explanation:

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