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elena-14-01-66 [18.8K]
3 years ago
13

Justin and his sister Stephanie are discussing

Mathematics
1 answer:
Solnce55 [7]3 years ago
5 0
84% -Justin
90% -Stephanie

Stephanie scored 6 points higher.
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Solve the system of equations using subtraction. Equation 1: 3x-2y= -2 Equation 2: 3x + y = 10 The solution is (? ,?) HURRY PLEA
Art [367]

Answer:

the solution is x = 2 and y = 4 OR (2,4)

Step-by-step explanation:

From the question

Equation 1: 3x-2y= -2

Equation 2: 3x + y = 10

Subtract equation 1 from equation 2, that is

3x + y = 10 - (3x -2y = -2)

You get

3x - 3x + y - (-2y) = 10 - (-2)

0 + y + 2y = 10 + 2

3y = 12

Divide both sides by 3

∴ y = 12/3

y = 4

Substitute the value of y into equation 2 to get x

3x + y = 10

3x + 4 = 10

Then,

3x = 10 - 4

3x = 6

Divide both sides by 3

∴ x = 6/3

x = 2

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Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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3 years ago
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