Answer:
The solution is x=1,y=2,z=3
Step-by-step explanation:
The given system of equations is ;\
2x−3y+4z=8...(1)
3x+4y−5z=−4...(2)
4x−5y+6z=12...(3)
Make x the subject in equation (1)
Put equation (4) into equation (2) and (3)
Multiply through by;
Expand;
Simplify;
Equation (4) in (3)
Put equation (6) into equation (5)
z=3
Put z=3 into equation (6)
y=2(3)-4=2
Put y=2 and z=3 into equation 4
The solution is x=1,y=2,z=3
Answer & Step-by-step explanation:
The formula is given in the middle. You are supposed to insert the value of x into this formula to find the value of y (like in the first given example):
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:Done
Answer:
Step-by-step explanation:
2{5x²-15+(-9xy²)}-(2y²+4x-xy²)+3x²
=2{5x²-15-9xy²}-(2y²+4x-xy²)+3x²
=10x²-30-18xy²-2y²-4x+xy²+3x²
=13x²-2y²-17xy²-4x-30
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
