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-BARSIC- [3]
3 years ago
12

What is the solution to the following system of equations? y=6x-11 ////// -2x-3y=-7

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
6 0
Y = 6x - 11
-2x - 3y = -7

Let's use substitution:

-2x - 3y = -7
-2x - 3(6x - 11) = -7
-2x - 18x + 33 = -7
-20x + 33 = -7
-20x = -40
x = 2

----

y = 6x - 11
y = 6(2) - 11
y = 12 - 11
y = 1

----

y = 6x - 11  =>  1 ? 6(2) - 11  =>  1 = 12 - 11  <--True
-2x - 3y = -7  =>  -2(2) - 3(1) ? -7  =>  -4 - 3 = -7  <--True

Answer:
x = 2
y = 1

Hope this helps!
Masja [62]3 years ago
4 0

Answer:

D. (2,1)

Step-by-step explanation:

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The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

10x^{2} -17x+3=0

so

a=10\\b=-17\\c=3

substitute in the formula

x=\frac{-(-17)(+/-)\sqrt{-17^{2}-4(10)(3)}} {2(10)}

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\frac{f(b)-f(a)}{b-a}

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4x^{2}=1

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This is a vertical parabola open upward

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The equation of a vertical parabola in vertex form is equal to

y=a(x-h)^2+k

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a is a coefficient

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(h,k)=(4,-1)

substitute

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3=a(2-4)^2-1

3=4a-1

4a=4

a=1

The quadratic equation is

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so

For y=0

(x-4)^2-1=0

Solve for x

(x-4)^2=1

take square root both sides

(x-4)=(+/-)1

x=4(+/-)1

x=4+1=5\\x=4-1=3

therefore

The x-intercepts are the points (3,0) and (5,0)

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