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4 years ago

How do you solve Y=x+1 x+2y=14 in substitution?

Mathematics

1 answer:

AleksAgata [21]4 years ago

6 0

Answer:

(4, 5 )

Step-by-step explanation:

Given the 2 equations

y = x + 1 → (1)

x + 2y = 14 → (2)

Substitute y = x + 1 into (2)

x + 2(x + 1) = 14 ← distribute and simplify left side

x + 2x + 2 = 14

3x + 2 = 14 ( subtract 2 from both sides )

3x = 12 ( divide both sides by 3 )

x = 4

Substitute x = 4 into (1) for corresponding value of y

y = 4 + 1 = 5

Solution is (4, 5 )

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4 years ago

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What must be the sides of an equilateral triangle so that it's area should be equal to the area of an isosceles triangle with ba

d1i1m1o1n [39]

Answer:

The sides of the equilateral triangle are 11.7 m.

Step-by-step explanation:

Let's find the area of the isosceles triangle:

$A_{i} = \frac{bh}{2}$

Where:

b: is the base = 12 m

h: is the height

We can find the height by using Pitagoras:

$x^{2} = \frac{b^{2}}{2} + h^{2}$

Where:

x is the hypotenuse = side of the triangle = 10 m

$h = \sqrt{x^{2} - \frac{b^{2}}{2}} = \sqrt{(10)^{2} - 6^{2}} = 8 m$

Then, the area is:

$A_{i} = \frac{12*8}{2} = 48 m^{2}$

Now, since the area of the isosceles triangle is equal to the area of the equilateral triangle:

$A_{e} = A_{i} = 48 m^{2}$

$A_{e} = \frac{bh}{2}$

The height of the equilateral triangle is given by:

$b^{2} = \frac{b^{2}}{2} + h^{2}$

$h = \sqrt{b^{2} - \frac{b^{2}}{2}} = \frac{b}{\sqrt{2}}$

Hence, the sides are:

$A_{e} = \frac{1}{2}b\frac{b}{\sqrt{2}} = \frac{b^{2}}{2\sqrt{2}}$

$b = \sqrt{A*2\sqrt{2}} = \sqrt{48*2\sqrt{2}} = 11.7 m$

Therefore, the sides of the equilateral triangle are 11.7 m.

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3 years ago

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