Q&A for How to Solve Systems of Algebraic Equations Containing Two Variables

The second step is to change R to 2R/2. (It's easier to subtract R/2 from 2R/2.) Realize that "R/2" means "one-half of an R." So "R - R/2" means "subtract half-an-R from a full R," which leaves you with half-an-R. So the equation really means that half-an-R equals 6. That means that a full R equals 12.

The substitution method often involves the least amount of work, but the elimination method is sometimes easier. It just depends on the equations involved.

The original cost was defined as Rs. 5,000.

Assuming you mean that no equation has been given, the expression cannot be "solved."

If you mean that there's one equation containing two variables, one of them appearing on both sides of the equation and the other appearing on only one side, move one of the variables by adding, subtracting, multiplying or dividing until one of the variables is on one side of the equation, and the other variable is on the other side. Then isolate either of the variables so that it stands alone without a coefficient or other accompanying numbers. You have now solved the equation for the isolated variable. You can do the same thing for the other variable, too, if you want. As for an equation in three variables, isolate one of the variables, and evaluate it in terms of the other two.

This equation has two unknowns. Therefore neither unknown can be evaluated except in terms of the other. So simplify the right side. Then solve for x in terms of y and/or for y in terms of x.

Presumably you have an equation in x and y. Choose conveniently low (positive and/or negative) values for the independent variable (x or y). Then calculate the corresponding values for the dependent variable. Place these values in the appropriate columns of the table.

The tangent of 20° is 0.364. h = (d + 14.37)(0.364) = 0.364d + 5.23. (Without knowing the value of d, this is as far as you can go.)

2x + 18 = 20. So 2x = 20 - 18 = 2, and x = 1. Then 3x = (3)(1) = 3.

Adding these two equations together, we get 9y = 22, so y = 22/9. Then 5x + 3(22/9) = 12. 5x + 66/9 = 12. 5x = 12 - 66/9 = 108/9 - 66/9 = 39/9. Divide both sides by 5, so that x = 39/45 = 13/15.

Use the elimination method. Add the two equations together: 3a = 1. So a = 1/3. Plug that a-value into either original equation: (1/3) + b = 0. So b = -1/3.

Assuming you mean that all three equations contain both unknowns, you could ignore one of the equations, because all you need is two. Use the usual steps in solving for two unknowns using two simultaneous equations. See Solve Systems of Algebraic Equations Containing Two Variables .

Add 30 to both sides: 5x = 40. Divide both sides by 5: x = 8.

Because that equation contains two variables, all you can do is solve for either of the variables in terms of the other. First subtract 15 from both sides of the equation so that 2x + 10y = 1005. Then 2x = 1005 - 10y, and x = (1005 - 10y) / 2. Similarly, 10y = 1005 - 2x, and y = (1005 - 2x) / 10. (To get a purely numerical value for a variable, you would need an equation containing only one variable, or if you have two variables, you would need a system of two equations involving those variables.)

See Solve Equations with Variables on Both Sides .

You need to have at least as many equations as there are variables, with each equation expressed in terms of one or more of those variables. Then use the elimination method or the substitution method to solve for the variables.