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Solve for x, y
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Solve for x, y (complex solution)
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y-kx=b
Consider the first equation. Subtract kx from both sides.
x^{2}+4y^{2}=4
Consider the second equation. Multiply both sides of the equation by 4.
y+\left(-k\right)x=b,x^{2}+4y^{2}=4
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.
y+\left(-k\right)x=b
Solve y+\left(-k\right)x=b for y by isolating y on the left hand side of the equal sign.
y=kx+b
Subtract \left(-k\right)x from both sides of the equation.
x^{2}+4\left(kx+b\right)^{2}=4
Substitute kx+b for y in the other equation, x^{2}+4y^{2}=4.
x^{2}+4\left(k^{2}x^{2}+2bkx+b^{2}\right)=4
Square kx+b.
x^{2}+4k^{2}x^{2}+8bkx+4b^{2}=4
Multiply 4 times k^{2}x^{2}+2bkx+b^{2}.
\left(4k^{2}+1\right)x^{2}+8bkx+4b^{2}=4
Add x^{2} to 4k^{2}x^{2}.
\left(4k^{2}+1\right)x^{2}+8bkx+4b^{2}-4=0
Subtract 4 from both sides of the equation.
x=\frac{-8bk±\sqrt{\left(8bk\right)^{2}-4\left(4k^{2}+1\right)\left(4b^{2}-4\right)}}{2\left(4k^{2}+1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+4k^{2} for a, 4\times 2kb for b, and -4+4b^{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8bk±\sqrt{64b^{2}k^{2}-4\left(4k^{2}+1\right)\left(4b^{2}-4\right)}}{2\left(4k^{2}+1\right)}
Square 4\times 2kb.
x=\frac{-8bk±\sqrt{64b^{2}k^{2}+\left(-16k^{2}-4\right)\left(4b^{2}-4\right)}}{2\left(4k^{2}+1\right)}
Multiply -4 times 1+4k^{2}.
x=\frac{-8bk±\sqrt{64b^{2}k^{2}-16\left(b^{2}-1\right)\left(4k^{2}+1\right)}}{2\left(4k^{2}+1\right)}
Multiply -4-16k^{2} times -4+4b^{2}.
x=\frac{-8bk±\sqrt{16+64k^{2}-16b^{2}}}{2\left(4k^{2}+1\right)}
Add 64k^{2}b^{2} to -16\left(1+4k^{2}\right)\left(b^{2}-1\right).
x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{2\left(4k^{2}+1\right)}
Take the square root of -16b^{2}+64k^{2}+16.
x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}
Multiply 2 times 1+4k^{2}.
x=\frac{-8bk+4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}
Now solve the equation x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2} when ± is plus. Add -8kb to 4\sqrt{-b^{2}+4k^{2}+1}.
x=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}
Divide -8bk+4\sqrt{-b^{2}+4k^{2}+1} by 2+8k^{2}.
x=\frac{-8bk-4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}
Now solve the equation x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2} when ± is minus. Subtract 4\sqrt{-b^{2}+4k^{2}+1} from -8kb.
x=-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}
Divide -8kb-4\sqrt{-b^{2}+4k^{2}+1} by 2+8k^{2}.
y=k\times \frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}+b
There are two solutions for x: \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} and -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}. Substitute \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} for x in the equation y=kx+b to find the corresponding solution for y that satisfies both equations.
y=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}k+b
Multiply k times \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}.
y=k\left(-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\right)+b
Now substitute -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} for x in the equation y=kx+b and solve to find the corresponding solution for y that satisfies both equations.
y=\left(-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\right)k+b
Multiply k times -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}.
y=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}k+b,x=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\text{ or }y=\left(-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\right)k+b,x=-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}
The system is now solved.