<p style='text-indent:20px;'>In the paper under study, we consider the following coupled non-degenerate Kirchhoff system</p><p style='text-indent:20px;'><disp-formula><label/><tex-math id="FE0"> \begin{document}$\begin{equation} \left \{ \begin{aligned} & y_{tt}-\mathtt{φ}\Big(\int_\Omega | \nabla y |^2\,dx\Big)\Delta y +\mathtt{α} \Delta \mathtt{θ} = 0, &{\rm{ in }}&\; \Omega \times (0, +\infty)\\ & \mathtt{θ}_t-\Delta \mathtt{θ}-\mathtt{β} \Delta y_t = 0, &{\rm{ in }}&\; \Omega \times (0, +\infty)\\ & y = \mathtt{θ} = 0,\; &{\rm{ on }}&\;\partial\Omega\times(0, +\infty)\\ & y(\cdot, 0) = y_0, \; y_t(\cdot, 0) = y_1,\;\mathtt{θ}(\cdot, 0) = \mathtt{θ}_0, \; \; &{\rm{ in }}&\; \Omega\\ \end{aligned} \right. \end{equation} \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$ \end{document}</tex-math></disp-formula></p><p style='text-indent:20px;'>where <inline-formula><tex-math id="M1">\begin{document}$ \Omega $\end{document}</tex-math></inline-formula> is a bounded open subset of <inline-formula><tex-math id="M2">\begin{document}$ \mathbb{R}^n $\end{document}</tex-math></inline-formula>, <inline-formula><tex-math id="M3">\begin{document}$ \mathtt{α} $\end{document}</tex-math></inline-formula> and <inline-formula><tex-math id="M4">\begin{document}$ \mathtt{β} $\end{document}</tex-math></inline-formula> be two nonzero real numbers with the same sign and <inline-formula><tex-math id="M5">\begin{document}$ \mathtt{φ} $\end{document}</tex-math></inline-formula> is given by <inline-formula><tex-math id="M6">\begin{document}$ \mathtt{φ}(s) = \mathfrak{m}_0+\mathfrak{m}_1s $\end{document}</tex-math></inline-formula> with some positive constants <inline-formula><tex-math id="M7">\begin{document}$ \mathfrak{m}_0 $\end{document}</tex-math></inline-formula> and <inline-formula><tex-math id="M8">\begin{document}$ \mathfrak{m}_1 $\end{document}</tex-math></inline-formula>. So we prove existence of solution and establish its exponential decay. The method used is based on multiplier technique and some integral inequalities due to Haraux and Komornik[<xref ref-type="bibr" rid="b5">5</xref>,<xref ref-type="bibr" rid="b8">8</xref>].</p>