Question

a(0.3−y)+1.1+2.4x= 0
(y−1.2)=−1.2(x−0.5)

Consider the system of equations above, where a is a constant. For which value of a is there exactly one (x, y) solution where y = 1.3y? Round your answer to the nearest tenth.

Answer 1

we can rewrite the second equation, as follows:
\[\begin{gathered}
y - 1.2 = - 1.2x + 6 \hfill \\
\hfill \\
y = - 1.2x + 7.2 \hfill \\
\end{gathered} \]

Answer 2

next I substitute such value of \(y\), into the first equation, namely I am applying the substitution method:
\[0.3\left( {a - \left( { - 1.2x + 7.2} \right)} \right) + 1.1 + 2.4x = 0\]
please simplify the left side

Answer 3

oops... sorry I have made a typo:
\[a\left( {0.3 - \left( { - 1.2x + 7.2} \right)} \right) + 1.1 + 2.4x = 0\]
please simplify

Answer 4

hint:
if we apply the properties of the algebra of real numbers, namely the distributive property of multiplication, we get:
\[1.2\left( {a + 2} \right)x = 6.9 - 0.3a\]

Answer 5

the condition:
\(y=1.3y\) means \(1.3y-y=0\), so we get:
\(0.3y=0\), whose solution is \(y=0\).
Next I substitute \(y=0\) into the second equation:
\((0-1.2)=-1.2(x-0.5)\)
from which I get:
\(-1.2=-1.2x+0.6\)
and \(x=1.5\)
finally I substitute the values \(x=1.5,y=0\) into the first equation, and I get:
\[a\left( {0.3 - 0} \right) + 1.1 + \left( {2.4 \cdot 1.5} \right) = 0\]
please solve for \(a\)

Answer 6

a=0?

because after you simplify you are left with a(5)=0

or at least that's what I got

Answer 7

I got this:
\[0.3a + 4.7 = 0\]
so, we have:
\[a = \frac{{ - 4.7}}{{0.3}} = ...?\]
please complete