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.

answer: b

step-by-step explanation:

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb