Solving Rational Equations Practice

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(2+1x=7\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(3+2x=11\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(5+2x=17\). Thus \(x=6\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(4+3x=19\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(7+2x=15\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(1+4x=13\). Thus \(x=3\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(6+5x=21\). Thus \(x=3\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(8+3x=20\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(9+4x=25\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(5+6x=23\). Thus \(x=3\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(2+7x=30\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(4+5x=29\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(3+8x=35\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(7+4x=31\). Thus \(x=6\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(6+7x=34\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(8+5x=33\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(9+6x=39\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(5+9x=41\). Thus \(x=4\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(7+8x=47\). Thus \(x=5\), which is allowed and verifies the original equation.

State x≠0, multiply every term by x, solve, and check.

Because \(x\ne0\), multiply every term by \(x\): \(11+6x=35\). Thus \(x=4\), which is allowed and verifies the original equation.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne1\). Cross multiplication gives \(5(x+2)=3(x-1)\), so \(x=\frac{-13}{2}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne2\). Cross multiplication gives \(7(x+3)=4(x-2)\), so \(x=\frac{-29}{3}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne3\). Cross multiplication gives \(5(x+1)=2(x-3)\), so \(x=\frac{-11}{3}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne2\). Cross multiplication gives \(8(x+4)=5(x-2)\), so \(x=-14\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne4\). Cross multiplication gives \(7(x+2)=3(x-4)\), so \(x=\frac{-13}{2}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne1\). Cross multiplication gives \(9(x+5)=7(x-1)\), so \(x=-26\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne5\). Cross multiplication gives \(9(x+3)=4(x-5)\), so \(x=\frac{-47}{5}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne2\). Cross multiplication gives \(11(x+6)=8(x-2)\), so \(x=\frac{-82}{3}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne6\). Cross multiplication gives \(12(x+4)=5(x-6)\), so \(x=\frac{-78}{7}\); substitution confirms it.

Use cross multiplication only because one fraction equals one fraction, then check the denominator.

The restriction is \(x\ne3\). Cross multiplication gives \(13(x+7)=9(x-3)\), so \(x=\frac{-59}{2}\); substitution confirms it.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=1\). Clearing the denominator produces only \(x=1\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=2\). Clearing the denominator produces only \(x=2\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=3\). Clearing the denominator produces only \(x=3\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=4\). Clearing the denominator produces only \(x=4\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=5\). Clearing the denominator produces only \(x=5\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=-1\). Clearing the denominator produces only \(x=-1\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=-2\). Clearing the denominator produces only \(x=-2\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=-3\). Clearing the denominator produces only \(x=-3\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=6\). Clearing the denominator produces only \(x=6\), so the sole candidate is extraneous and the equation has no solution.

Write the denominator restriction before clearing; reject a candidate that violates it.

The original domain excludes \(x=7\). Clearing the denominator produces only \(x=7\), so the sole candidate is extraneous and the equation has no solution.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne1,5\). Multiplying by \((x-1)(x-5)\) gives \((x-5)+(x-1)=0\), so \(x=3\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne2,8\). Multiplying by \((x-2)(x-8)\) gives \((x-8)+(x-2)=0\), so \(x=5\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne-1,7\). Multiplying by \((x+1)(x-7)\) gives \((x-7)+(x+1)=0\), so \(x=3\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne3,9\). Multiplying by \((x-3)(x-9)\) gives \((x-9)+(x-3)=0\), so \(x=6\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne-4,2\). Multiplying by \((x+4)(x-2)\) gives \((x-2)+(x+4)=0\), so \(x=-1\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne5,11\). Multiplying by \((x-5)(x-11)\) gives \((x-11)+(x-5)=0\), so \(x=8\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne-3,9\). Multiplying by \((x+3)(x-9)\) gives \((x-9)+(x+3)=0\), so \(x=3\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne4,12\). Multiplying by \((x-4)(x-12)\) gives \((x-12)+(x-4)=0\), so \(x=8\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne-5,7\). Multiplying by \((x+5)(x-7)\) gives \((x-7)+(x+5)=0\), so \(x=1\), an allowed value that verifies the original equation.

Use the factored LCD on every term, solve the resulting linear equation, and check both excluded values.

Restrictions are \(x\ne6,14\). Multiplying by \((x-6)(x-14)\) gives \((x-14)+(x-6)=0\), so \(x=10\), an allowed value that verifies the original equation.