14. 1.5, 10 <- Answer
15. 5,1 <- Answer
Proof 14
Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5
y = 10
Proof 15.
Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer: {x = 5 y = 1
X=6
16 plus 2 is 18 and divided by 3 is 6
<span>This means the length of the side of an area, a rectangle, for example, is always the same.
Hope this helps!</span>
The product of any number and its reciprocal is ' 1 '.
Call the number ' Q '.
Whatever the number is, its reciprocal is 1/Q .
Their product is
( Q ) x ( 1/Q ) = ( Q/Q ) = 1
I think it is 32, bc the value of x is 20 so your problem would look like this; 2 ( (20) - 4 )
