There are 5 ways to solve a system.
The most popular is to write both in standard notation then add the equations together. The easiest to explain is to use substitution. Solve one for one of the variables then substitute in the other equation.
The other ways to solve are to use graphing and find the intersection. Determinants and matrices are the other two ways.
If you don't know something in a math equation you can replace it with a variable and then solve it algebraically.
No matter what you're finding, use the equation V=πr2h. Then solve algebraically. V=volume h=height r=radius πr2=area of the base
Written algebraically, the equation is (2/A)=4.The first step is to move the variable A from the denominator. To do this, multiply both sides of the equation by A:A*(2/A)=4*AThe A in the denominator on the left is canceled out and we are left with:2=4*ATo solve for A divide both sides of the equation by 4:2/4=ATwo over four simplifies to one half, which equals A:1/2=A
That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
solve it
If you don't know something in a math equation you can replace it with a variable and then solve it algebraically.
Algebraically manipulate the equation until you have the indicated variable on one side of the equation and all of the other factors on the other side.
To rearrange the equation for acceleration, you start with the equation (a = \frac{v_f - v_i}{t}) where (a) is acceleration, (v_f) is final velocity, (v_i) is initial velocity, and (t) is time. You can rearrange it to solve for any of the variables by manipulating the equation algebraically. For example, to solve for final velocity, you rearrange the equation as (v_f = v_i + a \times t).
The first step not possible in solving an equation algebraically is not to provide an equation in the first place in which it appears to be so in this case.
It depends on if the item is a cylinder, block, or pyramid. You would replace the appropriate geometric equation variables and solve for the unknown algebraically.
You get the exact solution.
No matter what you're finding, use the equation V=πr2h. Then solve algebraically. V=volume h=height r=radius πr2=area of the base
Written algebraically, the equation is (2/A)=4.The first step is to move the variable A from the denominator. To do this, multiply both sides of the equation by A:A*(2/A)=4*AThe A in the denominator on the left is canceled out and we are left with:2=4*ATo solve for A divide both sides of the equation by 4:2/4=ATwo over four simplifies to one half, which equals A:1/2=A
Sure. You can always 'solve for' a variable, and if it happens to be the only variable in the equation, than that's how you solve the equation.
you don't answer an equation, you solve an equation
That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
True ~APEX