Solve {l}{x^2/4+y^2/3=1}{x=my-1}

$$\left. \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1 } \\ { x = m y - 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{2\left(3m\sqrt{m^{2}+1}+2\right)}{3m^{2}+4}\text{, }y=\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}$$

$$x=\frac{2\left(3m\sqrt{m^{2}+1}-2\right)}{3m^{2}+4}\text{, }y=\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y^{2}=12$$

Consider the second equation. Subtract my from both sides.

$$x-my=-1$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$x+\left(-m\right)y=-1,4y^{2}+3x^{2}=12$$

Solve x+\left(-m\right)y=-1 for x by isolating x on the left hand side of the equal sign.

$$x+\left(-m\right)y=-1$$

Subtract \left(-m\right)y from both sides of the equation.

$$x=my-1$$

Substitute my-1 for x in the other equation, 4y^{2}+3x^{2}=12.

$$4y^{2}+3\left(my-1\right)^{2}=12$$

Square my-1.

$$4y^{2}+3\left(m^{2}y^{2}+\left(-2m\right)y+1\right)=12$$

Multiply 3 times m^{2}y^{2}+\left(-2m\right)y+1.

$$4y^{2}+3m^{2}y^{2}+\left(-6m\right)y+3=12$$

Add 4y^{2} to 3m^{2}y^{2}.

$$\left(3m^{2}+4\right)y^{2}+\left(-6m\right)y+3=12$$

Subtract 12 from both sides of the equation.

$$\left(3m^{2}+4\right)y^{2}+\left(-6m\right)y-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4+3m^{2} for a, 3\left(-1\right)\times 2m for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6m\right)±\sqrt{\left(-6m\right)^{2}-4\left(3m^{2}+4\right)\left(-9\right)}}{2\left(3m^{2}+4\right)}$$

Square 3\left(-1\right)\times 2m.

$$y=\frac{-\left(-6m\right)±\sqrt{36m^{2}-4\left(3m^{2}+4\right)\left(-9\right)}}{2\left(3m^{2}+4\right)}$$

Multiply -4 times 4+3m^{2}.

$$y=\frac{-\left(-6m\right)±\sqrt{36m^{2}+\left(-12m^{2}-16\right)\left(-9\right)}}{2\left(3m^{2}+4\right)}$$

Multiply -16-12m^{2} times -9.

$$y=\frac{-\left(-6m\right)±\sqrt{36m^{2}+108m^{2}+144}}{2\left(3m^{2}+4\right)}$$

Add 36m^{2} to 144+108m^{2}.

$$y=\frac{-\left(-6m\right)±\sqrt{144m^{2}+144}}{2\left(3m^{2}+4\right)}$$

Take the square root of 144+144m^{2}.

$$y=\frac{-\left(-6m\right)±12\sqrt{m^{2}+1}}{2\left(3m^{2}+4\right)}$$

Multiply 2 times 4+3m^{2}.

$$y=\frac{6m±12\sqrt{m^{2}+1}}{6m^{2}+8}$$

Now solve the equation y=\frac{6m±12\sqrt{m^{2}+1}}{6m^{2}+8} when ± is plus. Add 6m to 12\sqrt{1+m^{2}}.

$$y=\frac{12\sqrt{m^{2}+1}+6m}{6m^{2}+8}$$

Divide 6m+12\sqrt{1+m^{2}} by 8+6m^{2}.

$$y=\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}$$

Now solve the equation y=\frac{6m±12\sqrt{m^{2}+1}}{6m^{2}+8} when ± is minus. Subtract 12\sqrt{1+m^{2}} from 6m.

$$y=\frac{-12\sqrt{m^{2}+1}+6m}{6m^{2}+8}$$

Divide 6m-12\sqrt{1+m^{2}} by 8+6m^{2}.

$$y=\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}$$

There are two solutions for y: \frac{3\left(m+2\sqrt{1+m^{2}}\right)}{4+3m^{2}} and \frac{3\left(m-2\sqrt{1+m^{2}}\right)}{4+3m^{2}}. Substitute \frac{3\left(m+2\sqrt{1+m^{2}}\right)}{4+3m^{2}} for y in the equation x=my-1 to find the corresponding solution for x that satisfies both equations.

$$x=m\times \frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}-1$$

Multiply m times \frac{3\left(m+2\sqrt{1+m^{2}}\right)}{4+3m^{2}}.

$$x=\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m-1$$

Add m\times \frac{3\left(m+2\sqrt{1+m^{2}}\right)}{4+3m^{2}} to -1.

$$x=-1+\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m$$

Now substitute \frac{3\left(m-2\sqrt{1+m^{2}}\right)}{4+3m^{2}} for y in the equation x=my-1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=m\times \frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}-1$$

Multiply m times \frac{3\left(m-2\sqrt{1+m^{2}}\right)}{4+3m^{2}}.

$$x=\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m-1$$

Add m\times \frac{3\left(m-2\sqrt{1+m^{2}}\right)}{4+3m^{2}} to -1.

$$x=-1+\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m$$

The system is now solved.

$$x=-1+\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m,y=\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}\text{ or }x=-1+\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}m,y=\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{2\left(3m\sqrt{m^{2}+1}+2\right)}{3m^{2}+4}\text{, }y=\frac{3\left(-2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}\text{; }x=\frac{2\left(3m\sqrt{m^{2}+1}-2\right)}{3m^{2}+4}\text{, }y=\frac{3\left(2\sqrt{m^{2}+1}+m\right)}{3m^{2}+4}\text{, }&m\neq -\frac{2\sqrt{3}i}{3}\text{ and }m\neq \frac{2\sqrt{3}i}{3}\\x=-\frac{5}{2}=-2.5\text{, }y=-\frac{3}{2m}\text{, }&m=-\frac{2\sqrt{3}i}{3}\text{ or }m=\frac{2\sqrt{3}i}{3}\end{matrix}\right.$$