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3 years ago

Solve the simultaneous equations.

Mathematics

1 answer:

user100 [1]3 years ago

6 0

Answer:

x=3/2, y=-1/2

Step-by-step explanation:

2x+4y =1

3x-5y =7

We will use elimination to solve these

Multiply the first equation by 3

3(2x+4y =1) 6x +12y =3

Multiply the second equation by -2 so that the x coefficients are the opposites of each other (6 and -6)

-2(3x-5y =7) -6x +10y =-14

Add the two new equations together

6x +12y =3

-6x +10y =-14

--------------------

0 + 22y = -11

Divide each side by 22

22y/22 = -11/22

y = -1/2

Now we can substitute y=-1/2 into the first equation and solve for x

2x + 4y =1

2x +4(-1/2) =1

2x -2 =1

Add 2 to each side

2x-2+2=1+2

2x=3

Divide by 2

2x/2 =3/2

x = 3/2

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3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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3 years ago

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V125BC [204]

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Step-by-step explanation:

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