By solving a linear equation, we conclude that you bought 4 buns and 5 cakes.
<h3>
How many of each did I buy?</h3>
We will define the two variables:
- x = number of buns bought.
- y = number of cakes bought.
We know that you spent exactly £4.35, then we only need to solve the linear equation:
x*0.40 + y*0.55 = 4.35
We need to solve that equation for x and y, such that the values of x and y can only be positive whole numbers.
We can rewrite the equation as:
y*0.55 = 4.35 - x*0.40
y = (4.35 - x*0.40)/0.55
Now we only need to evaluate it in different values of x, and see for which value of x, the outcome y is also a whole number.
We will see that for x = 4, we have:
y = (4.35 - 4*0.40)/0.55 = 5
So you bought 4 buns and 5 cakes.
If you want to learn more about linear equations:
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In all poe ate 2/3 of the left overs
Answer:
option D
2x − 7(−3x − 17) = 4
Step-by-step explanation:
Given in the question two equations
<h3>Equation 1</h3>
2x - 7y = 4
<h3>Equation 2</h3>
3x + y = -17
Rearranging equation 2 in terms of y
3x + y = -17
y = -17 - 3x
Put this value of y in Equation 1
2x - 7y = 4
2x - 7(-17 - 3x) = 4
So,
If you use the substitution method to solve the following system, new equation will be
<h3>2x - 7(- 3x - 17) = 4 </h3>
Answer:
5^3 y^5
125 y^5
Step-by-step explanation:
5y^3/(5y)^-2
Distribute the exponent in the denominator
5y^3/(5 ^-2 y^-2)
A negative exponent in the denominator brings it to the numerator
5y^3 5 ^2 y^2
Combine like terms
5 * 5^2 * y^3 5^2
We know that a^b * a^c = a^(b+c)
5^(1+2) * y^( 3+2)
5^3 y^5
125 y^5