Question

Solve the following systems of equations:


`1.& \left\{\begin{matrix} 5x-3y-z=1&\\ x+4y-6z=-1& \\ 2x+3y+4z=9& \end{matrix}\right.\\ \ \\ 2.& \left\{\begin{matrix} x+y=7& \\ x* y=12& \end{matrix}\right.\\ \ \\ 3.& \left\{\begin{matrix} x^2+y^2=169& \\ x+y=17& \end{matrix}\right.\\ \ \\ 4. & \left\{\begin{matrix} y^2-2y+1=x & \\ √(x)+y=5 & \end{matrix}\right. \end{align*}`

Answer 1

`\large\displaystyle\text{$\begin{gathered}\sf 1) \ \left\{\begin{matrix} 5x-3y-z=1&\\ x+4y-6z=-1& \\ 2x+3y+4z=9& \end{matrix}\right. \end{gathered}$}`

The coefficient matrix associated with the system is constructed and the columns and rows are reduced.

`\left ( \begin{array}ccc 5 & -3 &-1 &1\\ 1 & 4 & -6 &-1\\ 2 & 3 & 4 & 9 \end{array} \right ) \begin{array}{c} \xrightarrow[\textup{F}_3- 2\textup{F}_2]{\textup{F}_1 - 5\textup{F}_2} \end{array} \left ( \begin{array}c 0 & -23& 29 &6\\ 1 & 4& -6 &-1\\ 0 & -5 &16 & 11 \end{array} \right )\\ \ \\ \ \\ \begin{array}{c} \xrightarrow[]{23\textup{F}_3 - 5\textup{F}_1} \end{array} \left ( \begin{array}ccc 0 & -23& 29 &6\\ 1 & 4& -6 &-1\\ 0 & 0 & 223 & 223 \end{array} \right )`

So, z=1. Translating the last matrix to the associated system of equations, we have that x=1, y=1, since:

`\large\displaystyle\text{$\begin{gathered}\sf \begin{align*} y&=(6-29)/(-23)=(-23)/(-23)=1\\ x = -1+6-4=1 \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \bf{2) \ \left\{\begin{matrix} x+y=7& \\ x* y=12& \end{matrix}\right. } \end{gathered}$}`

Clear an unknown in the first equation and substitute the resulting expression in the second. Next, the quadratic equation is solved.

`\large\displaystyle\text{$\begin{gathered}\sf \bf{ y=7-x \quad \Longrightarrow \quad 12=x(7-x)=7x-x^2 } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \bf{ 0=x^2-7x+12=(x-4)(x-3); } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \bf{ x_1=4\qquad x_2=3 \Longrightarrow y_1=3 \qquad y_2=4 } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{3) \ \left\{\begin{matrix} x^2+y^2=169& \\ x+y=17& \end{matrix}\right. } \end{gathered}$}`

Clear an unknown in the first equation and substitute the resulting expression in the second. Next, the quadratic equation is solved.

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ y=17-x \quad \Longrightarrow \quad 169=x^2 +(17-x)^2=x^2+x^2 -34x+289 } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ 0=2x^2-34x+120=2(x^2-17+60)=2(x-12)(x-5); } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{ x_1=4\qquad x_2=3 \Longrightarrow y_1=3 \qquad y_2=4 } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \pmb{4 \ \left\{\begin{matrix} y^2-2y+1=x & \\ √(x)+y=5 & \end{matrix}\right. } \end{gathered}$}`

Substitute the expression that represents x into the second equation. Then both sides of the equation are squared and solved.

`\large\displaystyle\text{$\begin{gathered}\sf \pmb{ √(y^2-2y+1)+y=5\quad \Longrightarrow\quad \left ( √(y^2-2y+1) \right )^2=(5-y)^2 } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \pmb{ y^2-2y+1=y^2-10y+25\quad \Longrightarrow\quad 8y=24; } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \pmb{ y=6 \qquad x=4 } \end{gathered}$}`