How many solutions does the system of equations have? y=-2x+3 y = x² - 6x +3

Mathematics

1 answer:

Mashutka [201]2 years ago

6 0

Answer:

2 solutions

Step-by-step explanation:

y = - 2x + 3 → (1)

y = x² - 6x + 3 → (2)

substitute y = x² - 6x + 3 into (1)

x² - 6x + 3 = - 2x + 3 ( subtract - 2x + 3 from both sides )

x² - 4x = 0 ← factor out x from each term on the left side

x(x - 4) = 0

equate each factor to zero and solve for x

x = 0

x - 4 = 0 ⇒ x = 4

substitute these values into (1)

x = 0 : y = - 2(0) + 3 = 0 + 3 = 3 ⇒ (0, 3 )

x = 4 : y = - 2(4) + 3 = - 8 + 3 = - 5 ⇒ (4, - 5 )

the 2 solutions are (0, 3 ) and (4, - 5 )

You might be interested in

X and y simplifying problem. Please explain how to do thiese. I will gove 15 points

Strike441 [17]

$\dfrac 56 x + \dfrac25 y - \dfrac14 x + \dfrac3{10} y$

First, combine like terms:

$\left(\dfrac 56 - \dfrac14\right) x + \left(\dfrac25 + \dfrac3{10}\right)y$

Now just simplify the coefficients of x and y. We do this by converting each pair of fractions to ones with a common denominator. Then we can combine them easily.

$\dfrac56 - \dfrac14 = \dfrac56\cdot\dfrac22 - \dfrac14\cdot\dfrac33 = \dfrac{5\cdot2}{6\cdot2} - \dfrac{1\cdot3}{4\cdot3} = \dfrac{10}{12} - \dfrac3{12} = \dfrac{10-3}{12} = \dfrac7{12}$

$\dfrac25 + \dfrac3{10} = \dfrac25\cdot\dfrac22 + \dfrac3{10} = \dfrac{2\cdot2}{5\cdot2} + \dfrac3{10} = \dfrac4{10} + \dfrac3{10} = \dfrac{4+3}{10} = \dfrac7{10}$