First we assume this x-2y+3z=9 is eq.1 -x+3y = -4 is eq.2 2x-5y+5z=17 is eq.3 First I subtracted eq.2 and 3 from eq.1 x - 2y + 3z = 9 -(-x + 3y) =-(-4) -(2x - 5y + 5z =-17 So here, if you do it carefully, you can notice that the x and y’s will cancel and you’re left with -2z on LHS and -4 on RHS -2z = -4 2z = 4 z = 2 Then we can substitute the value of z into eq.1 So it becomes: x - 2y + 3z = 9 x - 2y + 3*2 = 9 x - 2y + 6 = 9 x - 2y = 9 - 6 x - 2y = 3 Mark the above equation as eq.4 Now we add eq.2 to eq.4 so that: x - 2y = 3 -x + 3y=-4 Now x will cancel out and y will be left on LHS and -1 will be left on the RHS So, y=-1 Now to your wish, you can substitute this value of y into eq.2 or eq.4, but I substituted it into eq.4 as x is on itself with no coefficients or signs in front of it So, we get: x - 2y = 3 x - 2*(-1) = 3 x + 2 = 3 x = 3-2=1 Therefore, the final answer is: (x,y,z) = (1, -1, 2)
By the way, when you were taking two of the three linear-equations to solve for a variable, all of the problems were on the small side & all of the problems were difficult to decipher (blurry!). 🤔
Kindly help on this question 🙏🙏 Find the set of values of a for which the system of equations x − 2y − 2z + 7 = 0 2x + (a − 9)y − 10z + 11 = 0 Given that a = −3, show that the system of equations has no solution 3x − 6y + 2az + 29 = 0
Shuffle the stand-alone constants to the right, and the variable terms (i.e. those with x, y, and z) to the left. Treat (a - 9) and 2*a as constant coefficients. Construct the matrix equation, in the form of A*X = B, where you'll have a square matrix of coefficients (A), multiplied by a column matrix (X) of x, y, and z, which equals a constant matrix (B) of the three stand-alone constants. For the system to have just one solution, the square matrix will have a non-zero determinant. For the system to have either no solution, or an infinite number of solutions, you'll need a determinant of zero for the square matrix. To find a determinant of a 2x2 matrix, it's the product along the positive (down-right) diagonal, minus the product along the negative (down-left) diagonal. For a 3x3 matrix, which we have, we have 3 products along positive diagonals to add up, and 3 products negative diagonals to subtract. See rule of Sarrus. This will generate a quadratic formula to solve for a, which is: a^2 - 2*a - 15 = 0 It has two solutions, a = -3, and a=+5. To show that a = -3 gives us no solution, apply this value of a, and look for a pair of equations with a pattern of coefficients that are proportional to each other. For us, this will be: 1*x - 2*y - 2*z = -7, and 3*x - 6*y - 6*z = -29 As you can see, the coefficients are all proportional to the corresponding coefficients in the other equation, but the stand-alone constants aren't. This means the two equations are contradictory, and there is no solution for x, y, and z, when a = -3. It's a few more steps to do the same for a=5, but it is also the case that when a=5, there are two contradictory equations.
But when solving simultaneous equations don't you eliminate like variables? e.g. x and 3x makes -2x as you subtract, 5 and 5 gives 0, 3y and 2y gives y, etc.
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First we assume this
x-2y+3z=9 is eq.1
-x+3y = -4 is eq.2
2x-5y+5z=17 is eq.3
First I subtracted eq.2 and 3 from eq.1
x - 2y + 3z = 9
-(-x + 3y) =-(-4)
-(2x - 5y + 5z =-17
So here, if you do it carefully, you can notice that the x and y’s will cancel and you’re left with -2z on LHS and -4 on RHS
-2z = -4
2z = 4
z = 2
Then we can substitute the value of z into eq.1
So it becomes:
x - 2y + 3z = 9
x - 2y + 3*2 = 9
x - 2y + 6 = 9
x - 2y = 9 - 6
x - 2y = 3
Mark the above equation as eq.4
Now we add eq.2 to eq.4 so that:
x - 2y = 3
-x + 3y=-4
Now x will cancel out and y will be left on LHS and -1 will be left on the RHS
So, y=-1
Now to your wish, you can substitute this value of y into eq.2 or eq.4, but I substituted it into eq.4 as x is on itself with no coefficients or signs in front of it
So, we get: x - 2y = 3
x - 2*(-1) = 3
x + 2 = 3
x = 3-2=1
Therefore, the final answer is:
(x,y,z) = (1, -1, 2)
Great
Thank you 🎉
Can this method work on four variable equations?
Thank you very much
By the way, when you were taking two of the three linear-equations to solve for a variable, all of the problems were on the small side & all of the problems were difficult to decipher (blurry!). 🤔
Thank you sir
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Kindly help on this question
🙏🙏
Find the set of values of a for which the system of equations
x − 2y − 2z + 7 = 0
2x + (a − 9)y − 10z + 11 = 0
Given that a = −3, show that the system of equations has no solution
3x − 6y + 2az + 29 = 0
Shuffle the stand-alone constants to the right, and the variable terms (i.e. those with x, y, and z) to the left. Treat (a - 9) and 2*a as constant coefficients.
Construct the matrix equation, in the form of A*X = B, where you'll have a square matrix of coefficients (A), multiplied by a column matrix (X) of x, y, and z, which equals a constant matrix (B) of the three stand-alone constants. For the system to have just one solution, the square matrix will have a non-zero determinant. For the system to have either no solution, or an infinite number of solutions, you'll need a determinant of zero for the square matrix.
To find a determinant of a 2x2 matrix, it's the product along the positive (down-right) diagonal, minus the product along the negative (down-left) diagonal. For a 3x3 matrix, which we have, we have 3 products along positive diagonals to add up, and 3 products negative diagonals to subtract. See rule of Sarrus.
This will generate a quadratic formula to solve for a, which is:
a^2 - 2*a - 15 = 0
It has two solutions, a = -3, and a=+5.
To show that a = -3 gives us no solution, apply this value of a, and look for a pair of equations with a pattern of coefficients that are proportional to each other. For us, this will be:
1*x - 2*y - 2*z = -7, and
3*x - 6*y - 6*z = -29
As you can see, the coefficients are all proportional to the corresponding coefficients in the other equation, but the stand-alone constants aren't. This means the two equations are contradictory, and there is no solution for x, y, and z, when a = -3. It's a few more steps to do the same for a=5, but it is also the case that when a=5, there are two contradictory equations.
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But when solving simultaneous equations don't you eliminate like variables? e.g. x and 3x makes -2x as you subtract, 5 and 5 gives 0, 3y and 2y gives y, etc.
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