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