System of Linear Equations entered :
[1] y - 2x/3 = -1
[2] y + x = 4
// To remove fractions, multiply equations by their respective LCD
Multiply equation [1] by 3
// Equations now take the shape:
[1] 3y - 2x = -3
[2] y + x = 4
Graphic Representation of the Equations :
-2x + 3y = -3 x + y = 4
Solve by Substitution :
// Solve equation [2] for the variable x
[2] x = -y + 4
// Plug this in for variable x in equation [1]
[1] 3y - 2•(-y +4) = -3
[1] 5y = 5
// Solve equation [1] for the variable y
[1] 5y = 5
[1] y = 1
// By now we know this much :
y = 1
x = -y+4
// Use the y value to solve for x
x = -(1)+4 = 3
I hope this help you
Answer:
Alaina
Step-by-step explanation:
Although I might be wrong on this I dont think I am as it is the closest to -1 while being the farthest from -2
Answer:
B. 5¹³
Step-by-step explanation:
(5²)⁴ * 5⁵
With parenthesis you multiply the exponents:
(5²)⁴ becomes 5⁸
Add exponents when multiplying base numbers are the same:
5⁸ * 5⁵ = 5⁽⁸ ⁺ ⁵⁾ = 5¹³
The system is:
i) <span>2x – 3y – 2z = 4
ii) </span><span>x + 3y + 2z = –7
</span>iii) <span>–4x – 4y – 2z = 10
the last equation can be simplified, by dividing by -2,
thus we have:
</span>i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
iii) 2x +2y +z = -5
The procedure to solve the system is as follows:
first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:
i) 2x – 3y – 2z = 4
iii) 2x +2y +z = -5
2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.
Equalize:
3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9
similarly, using i and ii, eliminate x:
i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
multiply the second equation by 2:
i) 2x – 3y – 2z = 4
ii) 2x + 6y + 4z = –14
thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:
3y+2z+4=-6y-4z-14
9y+6z=-18
So we get 2 equations with variables y and z:
a) 5y+3z=-9
b) 9y+6z=-18
now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.
Let's use elimination method, multiply the equation a by -2:
a) -10y-6z=18
b) 9y+6z=-18
------------------------ add the equations:
-10y+9y-6z+6z=18-18
-y=0
y=0,
thus :
9y+6z=-18
0+6z=-18
z=-3
Finally to find x, use any of the equations i, ii or iii:
<span>2x – 3y – 2z = 4
</span>
<span>2x – 3*0 – 2(-3) = 4
2x+6=4
2x=-2
x=-1
Solution: (x, y, z) = (-1, 0, -3 )
Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:
check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
Answer:
addition is commutative. 4+(6+(-2)=8 whiles
Step-by-step explanation:
4+6-2=8
