Solve the following system using elimination:
{15 y - 14 x = 15 | (equation 1)
{21 x - 20 y = -10 | (equation 2)
Swap equation 1 with equation 2:
{21 x - 20 y = -10 | (equation 1)
{-(14 x) + 15 y = 15 | (equation 2)
Add 2/3 × (equation 1) to equation 2:
{21 x - 20 y = -10 | (equation 1)
{0 x+(5 y)/3 = 25/3 | (equation 2)
Multiply equation 2 by 3/5:
{21 x - 20 y = -10 | (equation 1)
{0 x+y = 5 | (equation 2)
Add 20 × (equation 2) to equation 1:
{21 x+0 y = 90 | (equation 1)
{0 x+y = 5 | (equation 2)
Divide equation 1 by 21:
{x+0 y = 30/7 | (equation 1)
{0 x+y = 5 | (equation 2)
Collect results:
Answer: {x = 30/7, y = 5
Answer: what’s the problem?
You want to isolate one of the variables (x or y) so you can plug it into the other equation. The easiest one is isolating the 2nd equation.
3x² - 16x + 13 - y = 0 Add y on both sides
3x² - 16x + 13 = y
You can use this and plug it into the first equation
y - 12x + 15 = 3x²
(3x² - 16x + 13) - 12x + 15 = 3x²
3x² - 16x + 13 - 12x + 15 = 3x² Combine like terms
3x² - 28x + 28 = 3x² Subtract 3x² on both sides
-28x + 28 = 0 Add 28x on both sides
28 = 28x Divide 28 on both sides
1 = x
Now that you know x, you can plug it into either of the equation to find y
3(1)² - 16(1) + 13 - y = 0
3 - 16 + 13 - y = 0
-y = 0 Divide -1 on both sides
y = 0
x = 1, y = 0
It forms two acute angles so A
The <em><u>correct answer</u></em> is:
He can buy 3 packs of 10 or 30 single pieces.
Explanation:
He needs 30 pieces. Since they are sold in 10 packs, it would take 30/10 = 3 packs of 10.
Other than this, the pieces are sold as singles. This means he could also buy 30 single pieces.
