Given system of equations

\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)

\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)

\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)

This implies that

\(\displaystyle\triangle={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{10}\backslash{12}&{3}&-{5}\backslash{15}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)

since \(\displaystyle\triangle\ne{q}{0}\),there exits a unique solution for the given system.

\(\displaystyle\triangle_{{{x}{1}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{151}&{7}&-{10}\backslash{86}&{3}&-{5}\backslash{187}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)

\(\displaystyle\triangle_{{{x}{2}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&-{151}&-{10}\backslash{12}&{86}&-{5}\backslash{15}&{187}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)

\(\displaystyle\triangle_{{{x}{3}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{151}\backslash{12}&{3}&{86}\backslash{15}&-{9}&{187}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)

By cramer's rule

\(\displaystyle{x}_{{1}}={\frac{{\triangle{x}_{{1}}}}{{\triangle}}},{x}_{{2}}={\frac{{\triangle{x}_{{2}}}}{{\triangle}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{x}_{{3}}}}{{\triangle}}}\)

\(\displaystyle{x}_{{1}}={\frac{{{11490}}}{{{1149}}}},,{x}_{{2}}={\frac{{\triangle-{3447}}}{{{1149}}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{5745}}}{{{1149}}}}\)

\(\displaystyle{x}_{{1}}={10},{x}_{{2}}=-{3}{\quad\text{and}\quad}{x}_{{3}}={5}\)

Thus,the solution is \(\displaystyle{\left({x}_{{1}},{x}_{{2}},{x}_{{3}}\right)}={\left({10},-{3},{5}\right)}\)

\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)

\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)

\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)

This implies that

\(\displaystyle\triangle={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{10}\backslash{12}&{3}&-{5}\backslash{15}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)

since \(\displaystyle\triangle\ne{q}{0}\),there exits a unique solution for the given system.

\(\displaystyle\triangle_{{{x}{1}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{151}&{7}&-{10}\backslash{86}&{3}&-{5}\backslash{187}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)

\(\displaystyle\triangle_{{{x}{2}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&-{151}&-{10}\backslash{12}&{86}&-{5}\backslash{15}&{187}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)

\(\displaystyle\triangle_{{{x}{3}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{151}\backslash{12}&{3}&{86}\backslash{15}&-{9}&{187}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)

By cramer's rule

\(\displaystyle{x}_{{1}}={\frac{{\triangle{x}_{{1}}}}{{\triangle}}},{x}_{{2}}={\frac{{\triangle{x}_{{2}}}}{{\triangle}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{x}_{{3}}}}{{\triangle}}}\)

\(\displaystyle{x}_{{1}}={\frac{{{11490}}}{{{1149}}}},,{x}_{{2}}={\frac{{\triangle-{3447}}}{{{1149}}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{5745}}}{{{1149}}}}\)

\(\displaystyle{x}_{{1}}={10},{x}_{{2}}=-{3}{\quad\text{and}\quad}{x}_{{3}}={5}\)

Thus,the solution is \(\displaystyle{\left({x}_{{1}},{x}_{{2}},{x}_{{3}}\right)}={\left({10},-{3},{5}\right)}\)