Identify a possible first step using the elimination method to solve the system and then find the solution to the system. 3x - 5y = -2 2x + y = 3 Responses A Multiply first equation by -3 and second equation by 2, solution (1, -1).Multiply first equation by -3 and second equation by 2, solution (1, -1). B Multiply first equation by -2 and second equation by 3, solution (1, -1).Multiply first equation by -2 and second equation by 3, solution (1, -1). C Multiply first equation by -2 and second equation by 3, solution (1, 1).Multiply first equation by -2 and second equation by 3, solution (1, 1). D Multiply first equation by -3 and second equation by 2, solution (-1, 1)
Answer:
(C) Multiply first equation by -2 and second equation by 3, solution (1, 1)
Step-by-step explanation:
Simultaneous equations:Simultaneous equations are set of equations which possess a common solution. The equations can be solved by eliminating one of the unknowns by multiplying each of the equations in a way that a common coefficient is obtained in the unknown to be eliminated.
Given the simultaneous equations:
3x - 5y = -2
2x + y = 3
First step:
Multiply first equation by -2 and multiply second equation by 3,
-6x + 10y = 4
6x + 3y = 9
Second step:
Add the two equations together,
13y = 13
Divide both sides by 13
y = 1
Third step:
Put y = 1 in the first equation
3x - 5(1) = -2
3x - 5 = -2
3x = 5 - 2
3x = 3
Divide both sides by 3:
x = 1
solution (x,y) = (1,1)
Option C
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domain of fx=underroot x+2 is equal where {} denotes the fractional part
The domain of the function f(x) = {√(x + 2)} is x ≥ -2, where {} denotes the fractional part.
To find the domain of the function, we need to consider the values of x for which the expression inside the square root is defined. In this case, the expression x + 2 must be non-negative (since we cannot take the square root of a negative number). So we set x + 2 ≥ 0 and solve for x.
x + 2 ≥ 0
x ≥ -2
Therefore, the domain of the function is x ≥ -2. This means that the function is defined for all real numbers greater than or equal to -2.
In the second paragraph, we explain that the domain of the function f(x) = {√(x + 2)} is x ≥ -2. The function involves taking the square root of (x + 2) and then considering the fractional part of the result.
The expression inside the square root must be non-negative, which means x + 2 ≥ 0. Solving this inequality, we find x ≥ -2. Thus, any real number greater than or equal to -2 is included in the domain of the function.
For example, if we choose x = -2, the expression inside the square root becomes 0, and taking the square root of 0 gives us 0. The fractional part of 0 is also 0. As we move to x > -2, the square root expression becomes positive, resulting in non-zero fractional parts.
So, the function is defined and has a non-zero fractional part for all real numbers greater than -2. However, for x < -2, the expression inside the square root becomes negative, which is not defined in the real number system. Therefore, the domain of the function is x ≥ -2.
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