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Question

Question: How do you solve for y in \[4x + 7y = 28\] ?...

How do you solve for y in 4x+7y=284x + 7y = 28 ?

Explanation

Solution

Here in this given equation is a linear equation. Here we have to solve for one variable. To solve this equation for y by using arithmetic operation we can shift the x variable to RHS then solve the equation for y and on further simplification we get the required solution for the above equation.

Complete step-by-step solution:
The given equation is a linear equation. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. The general representation of the straight-line equation is y=mx+by = mx + b, it involves only a constant term and a first-order (linear) term, where m is the slope and b is the y-intercept. Occasionally, this equation is called a "linear equation of two variables," where y and x are the variables.
Consider the given equation
4x+7y=28\Rightarrow \,\,\,\,4x + 7y = 28
We have to shift the variable x and its coefficient to the RHS, by add -4x on both sides, then
4x+7y4x=284x\Rightarrow \,\,\,\,4x + 7y - 4x = 28 - 4x
On simplification we get
7y=284x\Rightarrow \,\,\,\,7y = 28 - 4x
To solve the equation for y, divide 7 by both sides, then
7y7=284x7\Rightarrow \,\,\,\,\dfrac{{7y}}{7} = \dfrac{{28 - 4x}}{7}
y=28747x\Rightarrow \,\,\,\,y = \dfrac{{28}}{7} - \dfrac{4}{7}x
y=447x\Rightarrow \,\,\,\,y = 4 - \dfrac{4}{7}x
Hence, the y value of the given linear equation 4x+7y=284x + 7y = 28 is y=5272xy = \dfrac{5}{2} - \dfrac{7}{2}x.

Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. We have 3 types of algebraic expressions namely monomial expression, binomial expression and polynomial expression. By using the tables of multiplication, we can solve the equation.