Which system of linear equations appears to have a solution of (-1,4)?
1 answer:
Answer:
The 3rd graph represents the point (-1, 4) as the solution to the system of linear equations.
The correct graph is also attached below.
Step-by-step explanation:
We know when two lines meet at a certain point, the point where they meet is called the point of intersection.
We also know that the point of intersection of two lines is termed as the solution of the system of linear equations.
We are given the point (-1, 4) as the solution of the system of linear equations.
It means the two lines meet at x = -1, and y = 4.
In other words,
(x, y) = (-1, 4) is called the point of intersection of the system of linear equations.
Now, if we closely check the graph of these two lines, we can easily figure out that the 3rd from the right side is the graph that represents the solution to the system of linear equations.
The 3rd graph indicates that the two lines meet at x = -1, and y = 4.
Hence, the point of intersection of 3rd graph is:
(x, y) = (-1, 4)
Therefore, the 3rd graph represents the point (-1, 4) as the solution to the system of linear equations.
The correct graph is also attached below.
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Answer:
Step-by-step explanation:
To solve these equations involving variables and exponents we need to follow these steps.
1) We need to find out the factor that is common in the equation.
2) After taking common, solve the equation. We can add or subtract only those values that have same bases.
1)
here we can see, both numbers are divisible by 2, so taking 2 common
It cannot be further simplified because both number donot have same bases.
3.
We can take 4n common
5. -12a -3
Here -3 cam be taken common
= -3(-12a/-3 -3/-3)
= -3(4a +1)
7.
here the smallest power of n is n^3 so, we can take n^3 common and both coefficients are divisible by 4 so taking 4n^3 common
9.
Here we cannot take k common, as k is not a multiple of 10. For taking common it should be divisible by each value in the equation. But each value s divisible by 5 so, taking 5 common
11.
Here we cannot take n common, as n is not a multiple of -60. For taking common it should be divisible by each value in the equation. But each value s divisible by 10 so, taking 10 common
13.
Here we cannot take n common, as n is not a multiple of 28. For taking common it should be divisible by each value in the equation. But each value s divisible by -4 so, taking -4 common
15.
Here we cannot take n common, as n is not a multiple of 18. For taking common it should be divisible by each value in the equation. But each value s divisible by 9 so, taking 9 common
17.