Mister Exam
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How to use it?
- Express {x} in terms of y where:
- 10*x-1*y=10
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- 15*x-17*y=19
- 9*x-8*y=6
- Equation:
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Equation sqrt(2+log2(x))=log2(x)
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Equation log(27)*x=1/3
- Equation 3*x^4+9*x^2-30=0
- Equation (4sin^2x-3)*sqrt(x^2-36pi^2)=o
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Identical expressions
- sqrt(x^ two +y^ two +4x-6y- twelve)+(x^ two +10x- twenty-seven)= eight -|y- nine |+|y- three |
- square root of (x squared plus y squared plus 4x minus 6y minus 12) plus (x squared plus 10x minus 27) equally 8 minus module of y minus 9| plus |y minus 3|
- square root of (x to the power of two plus y to the power of two plus 4x minus 6y minus twelve) plus (x to the power of two plus 10x minus twenty minus seven) equally eight minus module of y minus nine | plus |y minus three |
- √(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x2+y2+4x-6y-12)+(x2+10x-27)=8-|y-9|+|y-3|
- sqrtx2+y2+4x-6y-12+x2+10x-27=8-|y-9|+|y-3|
- sqrt(x²+y²+4x-6y-12)+(x²+10x-27)=8-|y-9|+|y-3|
- sqrt(x to the power of 2+y to the power of 2+4x-6y-12)+(x to the power of 2+10x-27)=8-|y-9|+|y-3|
- sqrtx^2+y^2+4x-6y-12+x^2+10x-27=8-|y-9|+|y-3|
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Similar expressions
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y+9|+|y-3|
- sqrt(x^2+y^2+4x-6y+12)+(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x^2-y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2-4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8+|y-9|+|y-3|
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|-|y-3|
- sqrt(x^2+y^2+4x-6y-12)-(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x+27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2+4x-6y-12)+(x^2-10x-27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2+4x+6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y+3|
Express x in terms of y where sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
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The solution
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__________________________ / 2 2 2 \/ x + y + 4*x - 6*y - 12 + x + 10*x - 27 = 8 - |y - 9| + |y - 3|
$$\left(\left(x^{2} + 10 x\right) - 27\right) + \sqrt{\left(- 6 y + \left(4 x + \left(x^{2} + y^{2}\right)\right)\right) - 12} = \left(8 - \left|{y - 9}\right|\right) + \left|{y - 3}\right|$$