Answer:
Check the solution below
Step-by-step explanation:
2) Given the equation
x +y =5... 1 and
x-y =3 ... 2
Add both equations
x+x = 5+3
2x = 8
x = 8/2
x = 4
Substitute x = 4 into 1:
From 1: x+y = 5
4+y= 5
y = 5-4
y = 1
3) Given
x+3y =15 ... 1
2x+7y=19 .... 2
From 2: x = 15-3y
Substitute into 2
2(15-3y)+7y = 19
30-6y+7y = 19
30+y = 19
y = 19-30
y = -11
Substitute y=-11 into x = 15-3y
x =15-3(-11)
x = 15+33
x = 48
The solution set is (48, -11)
4) given
x/2 +y/3 =0 and x+2y=1
From 1
(3x+2y)/6 = 0
3x+2y = 0.. 3
x+2y= 1... 4
From 4: x = 1-2y
Substutute
3(1-2y) +2y = 0
3-6y+2y = 0
3 -4y = 0
4y = 3
y = 3/4
Since x = 1-2y
x = 1-2(3/4)
x = 1-3/2
x= -1/2
The solution set is (-1/2, 3/4)
5) Given
5.x=1/2 and y =x +1 then solution is
We already know the vkue of x
Get y
y= x+1
y = 1/2 + 1
y = 3/2
Hence the solution set is (1/2, 3/2)
6) Given
3x +y =5 and x -3y =5
From 3; x = 5+3y
Substitute into 1;
3(5+3y)+y = 5
15+9y+y = 5
10y = 5-15
10y =-10
y = -1
Get x;
x = 5+3y
x = 5+3(-1)
x = 5-3
x = 2
Hence two solution set is (2,-1)
<span>The expression of the square root of 19x must be simplified when x is equal to 28. This is because possible factors of 28 can be seen to be 4 and 7, and 4 is a perfect square. This means it can be pulled outside of the square root when evaluated. The other options include only prime factors that could not be pulled out. (3,5), (3,7), (1,41)
28 simplifies as such:
Sqrt(19*28) = Sqrt(19*4*7) = 2*Sqrt(19*7) = 2*Sqrt(133).</span>
The image of circle A is missing, so i have attached it;
Answer:
x = 17
Step-by-step explanation:
In Δs EAB and EAD
We are told that;
∠BAE ≅ ∠DAE
AB = AD = AE
Now,
The Two triangles have two corresponding equal sides and the angles between them are equal, thus, we can say that the two triangles are congruent by SAS (Side Angle Side) postulate of congruence
By using the result of congruence, we can say that;
EB ≅ ED
We are given that;
EB = 3x - 24 and ED = x + 10
Thus,
3x - 24 = x + 10
∴ 3 x - 24 = x + 10
Add 24 to both sides to give;
3x = x + 34
Subtract x from both sides to give;
2x = 34
Divide both sides by 2 to give;
x = 17