With great difficulty because without an equality sign the given terms can't be considered to be an equation.
If the equation is 6 = -10 - p/5 then multiply all terms by 5 :-30 = -50 - p : p = -50 - 30 = -80
You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.
The question contains an algebraic expression but there is no equation to solve.
7r2 = 70r-175 Rearrange the equation and treat it as a quadratic equation: 7r2-70r+175 = 0 Divide all terms by 7: r2-10+25 = 0 Solve by factoring or using the quadratic equation formula or by completing the square: (r-5)(r-5) x = 5 and x also = 5 (they both have equal roots)
Without any equality sign and not knowing the plus or minus values of the given terms it can't be considered to be an equation.
2x + 5 - 5 = 2x. You can not solve for x because this is not an equation.
To solve for y in terms of x, divide both sides of the equation by 2: y = x/2.If x=2y then you have already solved for x.
What do you mean by "solve"? This is not an equation.
Add 5 to both sides of the equation to get rid of the - 5. -3x - 5 + 5 = -20 + 5 Solve and simplify. -3x = -15 Divide both sides of the equation by -3. x = 5
It combines like terms and results in the least amount if variables to solve for
You can't just solve one side of the equation. You always have to have both sides of the equal sign in order to manipulate the terms. If you have an equation like 10a + 5b=0, you can now solve for one of the terms (a or b), for example, b=10a/5 and a=-5b/10. Once you have solved for one term, you can plug it back into the equation and solve for real values. In this particular equation, a=-1/2b so by putting any value in for b (say 2), you get a definite value for a (-1) but there are an infinite number of such values.