If <em>x</em> = -1, you have
2(-1) + 3 cos(-1) + <em>e</em> ⁻¹ ≈ -0.0112136 < 0
and if <em>x</em> = 0, you have
2(0) + 3 cos(0) + <em>e</em> ⁰ = 4 > 0
The function <em>f(x)</em> = 2<em>x</em> + 3 cos(<em>x</em>) + <em>eˣ</em> is continuous over the real numbers, so the intermediate value theorem applies, and it says that there is some -1 < <em>c</em> < 0 such that <em>f(c)</em> = 0.
Step-by-step explanation:
4x+y=1-(1)
-6x+y=1-(2)
solve simultaneously
10x =0
X=0/10
X=0
solve for X in (1)
4(0)+y=1
0+y=1
y=1-0
y=1