Asystem of linear equations is shown below, where a and b are real numbers. 3x + 4y = a bx – 6y = 15 what values could a and b be for this system to have no solutions?
a. a = 6, b = –4.5
b .a = –10, b = –4.5
c. a = –6, b = –3
d. a = 10, b = –3

ANSWER

The equation has no solution if

and .

EXPLANATION

The first equation is

and the second equation is

We try solving this equation for B using the method of elimination.

We multiply equation (2) by 3 to obtain;

We also multiply equation (1) by B to obtain;

We now subtract equation (3) from  equation (4).

This will give us;

We factor on the right hand side to get;

We solve for y to obtain;

The above equation is defined for all real values except

This implies that

Remember that the system will have no solution if and only if the denominator is zero and the numerator is not equal to zero.

This means that,

When we substitute , then we will have

Hence the correct answer is option A

Note however that if both the numerator and denominator equal zero, then the system has infinitely many solutions.

∠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.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58