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3 years ago

Find the solution to the system of equations.y=2x+3 and y=-3x+3

Mathematics

1 answer:

iragen [17]3 years ago

6 0

Answer:

x = 0 and y = 3

Step-by-step explanation:

Using elimination method, we can solve for x and y in the simultaneous equation

y = 2x + 3 .................................eqn 1

y = -3x + 3 .................................eqn 2

multiply equation 1 by 3 and equation 2 by 2, then we will have

(y = 2x + 3) × 3

(y = -3x + 3) × 2

3y = 6x + 9 ...............................eqn 3

2y = -6x + 6 ................................eqn 4

Add eqn 3 and 4 together and we will eliminate x and have

5y = 15

Divide both sides by 5, and we will have

y = 3

To solve for x, substitute 3 for y in eqn 1

y = 2x + 3

3 = 2x + 3

collect like terms

3 - 3 = 2x

0 = 2x

Therefore x = 0

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3 years ago

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2 years ago

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3 years ago

An equation in the form ax2+bx+c=0 is solved by the quadratic formula. The solution to the equation is shown below.x=−7±√ "572"

jolli1 [7]

Answer:

B: a = 1, b= 7, c = -2

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0$

Given:

$x=\dfrac{-7\pm\sqrt{57}}{2}$

Comparing the terms of the given x-value with those of the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}=\dfrac{-7\pm\sqrt{57}}{2}$

Therefore:

$2a=2 \implies a=1$
$b = 7$
$b^2-4ac=57$

Using the found values of a and b to solve for c:

$\implies b^2-4ac=57$

$\implies (7)^2-4(1)c=57$

$\implies 49-4c=57$

$\implies -4c=57-49$

$\implies -4c=8$

$\implies c=-2$

In summary: a = 1, b = 7, c = -2

$\implies x^2+7x-2=0$

Therefore, option B is the correct solution.

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2 years ago

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A four digit number (not beginning with 0) can be represented by ABCD. There is one number such that ABCD=A^B*C^D, where A^B mea

shutvik [7]

The number (answer) is 2592.
Check your answer:
(substitute the numbers to see if they will equal 2592: 2^5 is 32 and 9^2 is 81)
Now subsitute it into A^B*C^D: 32*81 is 2592.

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3 years ago