Gdz Po — Algebre Klass Didakticheskii Material Zvavich

3⋅2x−3⋅5−2⋅x−2⋅4=13 center dot 2 x minus 3 center dot 5 minus 2 center dot x minus 2 center dot 4 equals 1 6x−15−2x−8=16 x minus 15 minus 2 x minus 8 equals 1

terms together and the constant numbers together on the left side. gdz po algebre klass didakticheskii material zvavich

A typical exercise from the 7th-grade Zvavic didactic material involves solving an equation with parentheses and distribution. Solve for 1. Distribute across parentheses Solve for Divide both sides by the coefficient

Introduces algebraic fractions, square roots, quadratic equations, and linear inequalities. gdz po algebre klass didakticheskii material zvavich

(6x−2x)+(-15−8)=1open paren 6 x minus 2 x close paren plus open paren negative 15 minus 8 close paren equals 1 4x−23=14 x minus 23 equals 1 Move the constant -23negative 23 to the right side by adding to both sides. 4x=1+234 x equals 1 plus 23 4x=244 x equals 24 4. Solve for Divide both sides by the coefficient of , which is x=244x equals 24 over 4 end-fraction x=6x equals 6 Final Result The value of that satisfies the equation