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snow_tiger [21]
3 years ago
5

Solve the following equation for x.

Mathematics
2 answers:
diamong [38]3 years ago
7 0

Answer:

b -10

hope this helps i think

mezya [45]3 years ago
7 0

Answer:

C. -5

Step-by-step explanation:

First, add 4 to both sides

9x = -49 + 4

Simplify

-49 + 4 into -45

so 9x = -45

Dividing both sides w/ 9

x = 45/9

Finally, simplify 45/9 into 5

x = -5

Sorry for a late answer...I hope this helps

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3 years ago
Which point is the solution to the following system of equations?
Naya [18.7K]

The point (3, 2) is the solution to given system of equations

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

Given that system of equations are:

x^2 + y^2 = 13    ------ eqn 1

2x - y = 4    ------- eqn 2

From eqn 2,

y = 2x - 4

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\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

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\begin{aligned}&x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4(5)(3)}}{2 \times 5}\\\\&x=\frac{16 \pm \sqrt{256-60}}{10}\\\\&x=\frac{16 \pm \sqrt{196}}{10}\end{aligned}

The discriminant b^2 - 4ac>0 so, there are two real roots.

\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}

Substitute for x = 0.2 and x = 3 in 2x - y = 4

<em><u>when x = 3</u></em>

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y = 2

<em><u>when x = 0.2</u></em>

2(0.2) - y = 4

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y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)

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