What is the solution (a, b) to this system of linear equations?

3a+6b=45

2a-2b=-12

(-27, 6)
(-1, 7)
(1, 7)
(27, -6)

c. (1,7)

Step-by-step explanation:

3a+6b=45     (i)

2a-2b=-12      (ii)

3a+6b=45 ⇒ a+2b= 15

and 2a-2b= - 12⇒ a-b= -6

a+2b= 15    (iii)  

a-b= -6        (iv)

we will get,

3b=21

On solving, the value of b is

b=7

a+2b= 15    

a+2(7)=15

a+14=15

a=15-14

a=1

Hence, the required values of (a, b) for given linear equations will be (1,7).

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a=1, b=7. (1, 7).

Step-by-step explanation:

3a+6b=45

2a-2b=-12

simplify 3a+6b=45 into a+2b=15

and 2a-2b=-12 into a-b=-6

a+2b=15

a-b=-6

a+2b=15

-(a-b)=-(-6)

a+2b=15

-a+b=6

3b=21

b=21/3

b=7

a+2(7)=15

a+14=15

a=15-14

a=1

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval