Since:
ax(2x + y - 5) = 6x^2 +3xy - 15x,
we should factor X out of the second equation to find out the value of a:
6x^2 + 3xy - 15x =
x(6x + 3y - 15),
Now that we factored X out, to find a, we need to find the other common factor:
(6x + 3y - 15) / (2x + y - 5),
which is 3.
Hence the value of a is 3
<h3>Given</h3>
z = 3
2y +z = 1
2x +3y +2z = 13
<h3>Find</h3>
x, y, z
<h3>Solution</h3>
... z = 3 is given
Substitute that into the second equation.
... 2y + 3 = 1
... 2y = -2 . . . . . subtract 3
... y = -1 . . . . . . . divide by the coefficient of y
Substitute these two solutions into the third equation.
... 2x +3(-1) +2(3) = 13
... 2x = 10 . . . . . . . . . . . . collect terms, subtract 3
... x = 5 . . . . . . . . . . . . . . divide by the coefficient of x
The variable values are: x = 5, y = -1, z = 3
Answer:
14 and 13
Step-by-step explanation:
14-12=2 which is less than 3
13-12=1 which is less than 3
