Solve the simultaneous equations
(x +3)(y+ 5)=24
(y +5)(z-7)=48
(z+ 7)(x+ 3)=32
The set of given equations is not easy to solve, but if there was a typo, and the equations were:
(x +3)(y+ 5)=24
(y +5)(z+7)=48
(z+ 7)(x+ 3)=32
Then the equation can be solved readily because of symmetry, namely
(x+3)=4, (y+5)=6, (z+7)=8
Can you confirm if there was a typo?
No typo
In a similar way, we can substitute
X=x+3
Y=y+5
Z=z+7, and
Z-14=z-7
The equations become:
XY=24 ...(1a)
Y(Z-14)=48 ...(2a)
ZX=32 ...(3a)
Divide (3a) by (1a) to get
Z/Y=32/24
Z=4Y/3 ...(4)
Substitute (4) in (2a) to get:
Y(4Y/3-14)=48
Solve the quadratic equation to get
Y=(±3sqrt(113)+21)/4
Substitute Y into 1a and 3a to get the remaining solutions of X, Y and Z.
Solve for x,y,z from X,Y,Z from the initial substitutions
X=x+3
Y=y+5
Z=z+7
I get
x=(11-3*sqrt(113)-11)/(sqrt(113)+7) or
x=-(3*sqrt(113)+11)/(sqrt(113)-7)
...
To solve the simultaneous equations, we can use the method of substitution or elimination. Let's use the method of substitution.
Step 1: Solve the first equation for one variable in terms of the other variable(s).
From the first equation, we have:
(x + 3)(y + 5) = 24
Expanding the equation, we get:
xy + 5x + 3y + 15 = 24
Rearranging terms, we have:
xy + 5x + 3y = 9
Step 2: Solve the second equation for one variable in terms of the other variable(s).
From the second equation, we have:
(y + 5)(z - 7) = 48
Expanding the equation, we get:
yz - 7y + 5z - 35 = 48
Rearranging terms, we have:
yz + 5z - 7y = 83
Step 3: Solve the third equation for one variable in terms of the other variable(s).
From the third equation, we have:
(z + 7)(x + 3) = 32
Expanding the equation, we get:
xz + 7x + 3z + 21 = 32
Rearranging terms, we have:
xz + 7x + 3z = 11
Step 4: Now we have a system of three linear equations:
xy + 5x + 3y = 9 ---(Equation 1)
yz + 5z - 7y = 83 ---(Equation 2)
xz + 7x + 3z = 11 ---(Equation 3)
We can solve this system of equations by substitution or elimination method. Let's use the elimination method.
Step 5: Multiply Equation 1 by 3, Equation 2 by -5, and Equation 3 by -7 to eliminate the y and z terms.
3(xy + 5x + 3y = 9) => 3xy + 15x + 9y = 27 ---(Equation 4)
-5(yz + 5z - 7y = 83) => -5yz - 25z + 35y = -415 ---(Equation 5)
-7(xz + 7x + 3z = 11) => -7xz - 49x - 21z = -77 ---(Equation 6)
Step 6: Add the modified equations (Equations 4, 5, and 6) to eliminate the y and z terms.
When we add the three equations, we get:
3xy + 15x + 9y - 5yz - 25z + 35y - 7xz - 49x - 21z = -165
Simplifying the equation, we have:
3xy - 5yz - 7xz + (15x + 9y + 35y - 25z - 49x - 21z) = -165
Combining like terms, we get:
3xy - 5yz - 7xz - 34x + 44y - 46z = -165 ---(Equation 7)
Step 7: Substitute the value of x in Equation 7 obtained from Equation 1.
From Equation 1, we have:
xy + 5x + 3y = 9
Simplifying the equation, we get:
xy + 5x = 9 - 3y
Solving for x, we find:
x = (9 - 3y) / (y + 5)
Step 8: Substitute the value of x in Equation 7 obtained from Equation 1.
Substituting the value of x in Equation 7, we get:
3[(9 - 3y) / (y + 5)]y - 5yz - 7[(9 - 3y) / (y + 5)]z - 34[(9 - 3y) / (y + 5)] + 44y - 46z = -165
Simplifying the equation, we get:
27y - 9y^2 - 5yz - 63z + 21y - 9yz - 34(9 - 3y) + 184y - 46z = -165(y + 5)
Simplifying further, we get a quadratic equation in terms of y and z:
9y^2 + 40yz - 46y - 13z - 306 = 0 ---(Equation 8)
Step 9: Solve Equation 8 to find the values of y and z.
To solve this quadratic equation, you can use the quadratic formula or factorization method. Once you find the values of y and z, substitute them into Equation 1 or Equation 2 or Equation 3 to find the value of x.
Note: The solution to this system of equations may involve complex numbers since the resulting quadratic equation may not have real roots.