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3 years ago

6

Determine whether each ordered pair is a solution of the given linear equation.

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

(2,1) is the only solution to the linear equation

Step-by-step explanation:

to solve, just plug in the values given into the orignal expression. remember, the answer on the left has to be equal to the answer on the right.

(2,1)

4(2) + 3(1) = 11 --> 8 + 3 = 11 --> 11 = 11 (2,1) is a solution to the linear equation

(6,0)

4(6) + 3(0) = 11 --> 24 + 0 = 11 --> 24 ≠ 11 (6,0) is not a solution to the linear equation

(0,1)

4(0) + 3(1) = 11 --> 0 + 3 = 11 --> 3 ≠ 11 (0,1) is not a solution to the linear equation

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3 years ago

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Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6

ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

- For $y=0$

$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

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3 years ago

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