To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.
Add 5y to both sides of the equation.
Divide both sides by 2.
Multiply times 5y - 13.
Substitute for x in the other equation, 3x + 4y = 15.
Multiply 3 times .
Add to 4y.
Add to both sides of the equation.
Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.
Substitute 3 for y in . Because the resulting equation contains only one variable, you can solve for x directly.
Multiply 5/2 times 3.
Add to by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.
The system is now solved. The value of x & y will be 1 & 3 respectively.
Divide the 90 stamps by 45: The result is 2; it's equal to 1% of the total number of stamps. You then multiply it by 100 to get the full amount, which is 200 stamps.
Since he has 3 part chocolate all he has 18 cups left. You can keep adding 3 till you get 18 and how many ever groups you have with 3 (which is 5 groups) is the answer.