Determine the values of a,c,b for the quadratic equation x²-8x+15

Mathematics

1 answer:

Zina [86]3 years ago

6 0

Answer: a= 1 b=-8 c=15

Step-by-step explanation:

A is the number before $x^{2}$

B is the number (including the sign) before X

c is the number without a variable

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Solve the system of linear equations by substituting. Check your solution

Ghella [55]

QUESTION 4

The given system of linear equations are,

$y = 5x - 7...(1)$

and

$- 3x - 2y = - 12...(2)$

We put equation (1) into equation (2) to obtain,

$- 3x - 2(5x - 7) = - 12$

We expand to get,

$- 3x - 10x + 14 = - 12$

Group like terms to get,

$- 3x - 10x = - 12 - 14$

Simplify to get,

$- 13x = - 26$

Divide both sides by -13 to get,

$x = 2$

Put the value of x into equation (1) to get,

$y = 5(2) - 7$

$y = 10 - 7 = 3$

The solution is

$x = 2 \: \: and \: \: = 3$

QUESTION 5

The given system of linear equations are

$-4x+y=6....(1)$
and

$-5x-y=21...(2)$

We make y the subject in equation (1) to get,

$y = 4x + 6....(3)$

Put equation (3) into equation (2) to get,

$- 5x - (4x + 6) = 21$

We expand to get,

$- 5x - 4x - 6 =21$

We group like terms to get,

$- 5x - 4x = 21 + 6$

This implies that,

$- 9x = 27$

We divide both sides by -9 to get,

$x = - 3$

Put the value of x in to equation 3 to get,

$y = 4( - 3) + 6$

$y = - 12 + 6 = - 6$

Therefore the solution is

$x = - 3 \: and \: y = - 6$

QUESTION 6

The given equations are,

$-7x-2y=-13...(1)$

and

$x-2y=-13...(2)$

Make x the subject in equation (2) to get,

$x = 2y - 13...(3)$

Put equation (3) into equation (1) to get,
$- 7(2y - 13) - 2y = - 13$

We expand to get,

$- 14y + 91 - 2y = - 13$

We group like terms to obtain,

$- 14y - 2y = - 13 - 91$

.Simplify to get,

$- 16y = - 104$

Divide through by -16 to get,

$y = \frac{13}{2}$

Put the value of y into equation (3) to get,

$x = 2( \frac{13}{2} ) - 13$

$x = 13 - 13 = 0$

Therefore

$x = 0 \: and \: y = \frac{13}{2}$