Solve the system of equations.

2 answers:

7 0

2x - 5y = 3
x - 3y = 1...multiply by -2
---------------
2x - 5y = 3
-2x + 6y = -2 (result of multiplying by -2)
--------------add
y = 1

x - 3y = 1
x - 3(1) = 1
x - 3 = 1
x = 1 + 3
x = 4

solution is (4,1)

BartSMP [9]3 years ago

6 0

Answer: First Option is correct.

Step-by-step explanation:

Since we have given that

$2x-5y = 3\\\\ x -3y = 1$

Now, we will use "Substitution Method" to find the values of x and y:

$x=1+3y\ -(3)$

Put the value of x in first equation i.e.

$2x-5y = 3$

$2(1+3y)-5y=3\\\\2+6y-5y=3\\\\2+y=3\\\\y=3-2\\\\y=1$

Now, we use the value of y in equation (3),

$x=1+3y\\\\x=1+3\times1\\\\x=1+3\\\\x=4$

Hence, values of (x,y) is (4,1).

Therefore, First Option is correct.

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A parabola is given by the equation y = x2 + 4x + 4.

notsponge [240]

Answer:

the vertex of the parabola is:(-2,0)

the focus of the parabola is:(-2, $\frac{1}{4}$ )

the directrix of the parabola is:y= $\frac{-1}{4}$

Step-by-step explanation:

we know that for any general equation of the parabola of the type $y=ax^{2} +b x+c$ the vertex of the parabola is given by (h,k)

where $h=\frac{-b}{2 a}$ and $k=\frac{4 a c-b^{2} }{4a}$

therefore by the given data we have h=-2 and k=0

hence vertex=(-2,0)

the general equation of the parabola of the type $4 a (y-h)=(x-k)^2$ ; the parabola symmetric around the y-axis has the focus from the centre i.e. the vertex (h,k) at a distance a as (h,k+a) and the directrix is given by y=k-a