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3 years ago

Solve these pair of simultaneous equations y=x^2-x-1 and y=2-3x

2 answers:

7 0

Y = x^2 - x - 1
y = 2 - 3x

2 - 3x = x^2 - x - 1
x^2 - x - 1 - 2 + 3x = 0
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0

x + 3 = 0 y = 2 - 3x
x = -3 y = 2 - 3(-3)
y = 2 + 9
y = 11

x - 1 = 0 y = 2 - 3x
x = 1 y = 2 - 3(1)
y = 2 - 3
y = -1

so ur solutions are : (-3,11) or (1,-1)

Gnom [1K]3 years ago

3 0

Compare both sides,
x² - x - 1 = 2 - 3x
x² + 2x - 3 = 0
x² -x + 3x - 3 = 0
x(x - 1) 3(x - 1) = 0
(x - 1)(x+3) = 0
x = 1 or - 3

In short, Your Answers would be 1 & -3

Hope this helps!

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3 years ago

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

4 0

3 years ago

Describe and correct the error made in solving the system of equations by elimination

lesya692 [45]

The mistake was when she multiplied the 2nd equation by 5, she forgot to also multiply that last number 10 by 5 as well.

correction :
5x + 3y - z = 15
-x + 2y + 3z = 10...multiply by 5
---------------------
5x + 3y - z = 15
-5x + 10y + 15z = 50 (result of multiplying by 5)
--------------------add
13y + 14z = 65

6 0

3 years ago