Step 1

Given the system of equations

2x-y=6...(1)

\(\displaystyle{x}^{{{2}}}+{y}={9}\)...(2)

To solve the equation:

Step 2

We use the substitution method to solve, from equation (1),

y=2x-6...(3)

Now, substitute in equation (2) and solve for x,

\(\displaystyle{x}^{{{2}}}+{\left({2}{x}-{6}\right)}={9}\)

\(\displaystyle{x}^{{{2}}}+{2}{x}-{6}={9}\)

\(\displaystyle{x}^{{{2}}}+{2}{x}-{15}={0}\)

\(\displaystyle{x}^{{{2}}}+{5}{x}-{3}{x}-{15}={0}\)

x(x+5)-3(x+15)=0

(x-3)(x+5)=0

\(\displaystyle\therefore{x}={3},-{5}\)

Step 3

Find the value of y corresponding to x,

Substitute the value of x as 3 in y,

y=2(3)-6=6-6=0

Substitute the value of x as -5 in y,

y=2(-5)-6=-10-6=-16

Step 4

Answer:

Thus, the solution of the given system is (3,0) and (-5,-16).

Given the system of equations

2x-y=6...(1)

\(\displaystyle{x}^{{{2}}}+{y}={9}\)...(2)

To solve the equation:

Step 2

We use the substitution method to solve, from equation (1),

y=2x-6...(3)

Now, substitute in equation (2) and solve for x,

\(\displaystyle{x}^{{{2}}}+{\left({2}{x}-{6}\right)}={9}\)

\(\displaystyle{x}^{{{2}}}+{2}{x}-{6}={9}\)

\(\displaystyle{x}^{{{2}}}+{2}{x}-{15}={0}\)

\(\displaystyle{x}^{{{2}}}+{5}{x}-{3}{x}-{15}={0}\)

x(x+5)-3(x+15)=0

(x-3)(x+5)=0

\(\displaystyle\therefore{x}={3},-{5}\)

Step 3

Find the value of y corresponding to x,

Substitute the value of x as 3 in y,

y=2(3)-6=6-6=0

Substitute the value of x as -5 in y,

y=2(-5)-6=-10-6=-16

Step 4

Answer:

Thus, the solution of the given system is (3,0) and (-5,-16).