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shepuryov [24]
4 years ago
10

OnSolve x + 11 = 0.​

Mathematics
2 answers:
r-ruslan [8.4K]4 years ago
5 0

Answer: x = -11

a negative 11 negates a positive 11 and brings you to 0

Step-by-step explanation:

forsale [732]4 years ago
4 0

The answer is Simplifying

x(x + 11) = 0

Reorder the terms:

x(11 + x) = 0

(11 * x + x * x) = 0

(11x + x2) = 0

Solving

11x + x2 = 0

Solving for variable 'x'.

Factor out the Greatest Common Factor (GCF), 'x'.

x(11 + x) = 0

Set the factor 'x' equal to zero and attempt to solve:

Simplifying

x = 0

Solving

x = 0

Move all terms containing x to the left, all other terms to the right.

Simplifying

x = 0

Set the factor '(11 + x)' equal to zero and attempt to solve:

Simplifying

11 + x = 0

Solving

11 + x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-11' to each side of the equation.

11 + -11 + x = 0 + -11

Combine like terms: 11 + -11 = 0

0 + x = 0 + -11

x = 0 + -11

Combine like terms: 0 + -11 = -11

x = -11

Simplifying

x = -11

Solution

x = {0, -11}

Step-by-step explanation:

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x(x + 7) + 8(x + 7)

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What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

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