Answer:
1 Use Difference of Squares:
a^2-b^2=(a+b)(a−b).
10(x^2-4^2)
2 Simplify 4^2 to 16
10(x^2 −16)
3 Expand by distributing terms.
10x^2 −160
Step-by-step explanation:
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§ALEX§
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:
3<5
Step-by-step explanation: dont see a picture but i have an example
When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
Simply substitute the point into the given equation; as there is one variable, you only need one point and so one is given. When done so, you shall get b=4
