Question

A) a+b=2 4a-3b=1 B)40m+3p=-10 20m-7p=-5

Answer 1

Ответ:

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Answer 2

Let's solve the given systems of equations:

A)

1. a + b = 2
2. 4a - 3b = 1

To solve this system, we can use the method of substitution or elimination. Let's use the elimination method:

Multiplying equation 1 by 3, we get: 3a + 3b = 6 (equation 3)

Now, we can subtract equation 2 from equation 3: (3a + 3b) - (4a - 3b) = 6 - 1 3a + 3b - 4a + 3b = 5 -a + 6b = 5 (equation 4)

Multiplying equation 4 by -1, we get: a - 6b = -5 (equation 5)

Now, we can add equation 1 to equation 5: (a + b) + (a - 6b) = 2 + (-5) 2a - 5b = -3 (equation 6)

Now we have a new system of equations:

1. a + b = 2
2. 2a - 5b = -3

We can solve this system using any method, let's use the elimination method again:

Multiplying equation 1 by 2, we get: 2a + 2b = 4 (equation 7)

Now, we can subtract equation 7 from equation 6: (2a - 5b) - (2a + 2b) = -3 - 4 2a - 5b - 2a - 2b = -7 -7b = -7 b = 1

Substituting the value of b into equation 1: a + 1 = 2 a = 1

So, the solution to the system of equations A) is: a = 1 and b = 1.

B)

1. 40m + 3p = -10
2. 20m - 7p = -5

We can solve this system using the elimination method:

Multiplying equation 1 by 7 and equation 2 by 3, we get: 280m + 21p = -70 (equation 3) 60m - 21p = -15 (equation 4)

Now, we can add equation 3 to equation 4: (280m + 21p) + (60m - 21p) = -70 + (-15) 340m = -85 m = -85/340 m = -1/4

Substituting the value of m into equation 1: 40(-1/4) + 3p = -10 -10 + 3p = -10 3p = 0 p = 0

So, the solution to the system of equations B) is: m = -1/4 and p = 0.