Use a General Strategy to Solve Linear Equations

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

Example. Solve: $-6(x+3) = 24$.

Example. Solve: $-(y+9) = 8$.

Simplify each side of the equation as much as possible by distributing: $-(y+9) = 8$ becomes $-y - 9 = 8$. The only $y$ term is on the left side, so all variable terms are already on the left side of the equation. Add $9$ to both sides to get all constant terms on the right side, and simplify: $-y = 17$. Rewrite $-y$ as $-1y$, then make the coefficient of the variable term equal to $1$ by dividing both sides by $-1$: $y = -17$.

Check: let $y = -17$. Then $-(-17+9) \overset{?}{=} 8$, so $-(-8) \overset{?}{=} 8$, and $8 = 8$. ✓

Example. Solve: $5(a-3)+5 = -10$.

Distribute: $5a - 15 + 5 = -10$. Combine like terms: $5a - 10 = -10$. The only $a$ term is on the left side, so all variable terms are on one side of the equation. Add $10$ to both sides to get all constant terms on the other side, and simplify: $5a = 0$. Make the coefficient of the variable term equal to $1$ by dividing both sides by $5$: $a = 0$.

Check: let $a=0$. Then $5(0-3)+5 \overset{?}{=} -10$, so $5(-3)+5 \overset{?}{=} -10$, so $-15+5 \overset{?}{=} -10$, and $-10 = -10$. ✓

Example. Solve: $\tfrac{2}{3}(6m - 3) = 8 - m$.

Distribute: $4m - 2 = 8 - m$. Add $m$ to both sides to get the variables only on the left: $5m - 2 = 8$. Add $2$ to both sides to get constants only on the right: $5m = 10$. Divide by $5$: $m = 2$.

Check: let $m = 2$. Then $\tfrac{2}{3}(6\cdot 2 - 3) \overset{?}{=} 8-2$, so $\tfrac{2}{3}(9) \overset{?}{=} 6$, and $6 = 6$. ✓

Example. Solve: $8 - 2(3y+5) = 0$.

Simplify — use the Distributive Property: $8 - 6y - 10 = 0$. Combine like terms: $-6y - 2 = 0$. Add $2$ to both sides to collect constants on the right: $-6y = 2$. Divide both sides by $-6$: $y = -\tfrac{1}{3}$.

Check: let $y = -\tfrac{1}{3}$. Then $8 - 2\left[3\left(-\tfrac{1}{3}\right)+5\right] \overset{?}{=} 0$, so $8 - 2(-1+5) \overset{?}{=} 0$, so $8 - 2(4) \overset{?}{=} 0$, so $8 - 8 \overset{?}{=} 0$, and $0 = 0$. ✓

Example. Solve: $4(x-1) - 2 = 5(2x+3) + 6$.

Distribute: $4x - 4 - 2 = 10x + 15 + 6$. Combine like terms: $4x - 6 = 10x + 21$. Since $10 > 4$, subtract $4x$ from both sides to get the variables only on the right side: $-6 = 6x + 21$. Subtract $21$ to get the constants on the left: $-27 = 6x$. Divide by $6$: $-\tfrac{9}{2} = x$.

Check: let $x = -\tfrac{9}{2}$. Substituting back into both sides of the original equation gives $-24 = -24$. ✓

Example. Solve: $10[3 - 8(2s-5)] = 15(40-5s)$.

Simplify from the innermost parentheses first: $10[3 - 16s + 40] = 15(40-5s)$. Combine like terms in the brackets: $10[43 - 16s] = 15(40-5s)$. Distribute: $430 - 160s = 600 - 75s$. Add $160s$ to get the $s$’s to the right: $430 = 600 + 85s$. Subtract $600$ to get the constants to the left: $-170 = 85s$. Divide: $-2 = s$.

Check: let $s=-2$. Substituting back into both sides gives $750 = 750$. ✓

Example. Solve: $0.36(100n+5) = 0.6(30n+15)$.

Distribute: $36n + 1.8 = 18n + 9$. Subtract $18n$ to get the variables to the left: $18n + 1.8 = 9$. Subtract $1.8$ to get the constants to the right: $18n = 7.2$. Divide by $18$: $n = 0.4$.

Check: let $n = 0.4$. Substituting back into both sides gives $16.2 = 16.2$. ✓

Classify Equations

Consider the equation we solved earlier, $7x+8=-13$. The solution we found was $x=-3$. This means the equation $7x+8=-13$ is true when we replace the variable, $x$, with the value $-3$. We showed this when we checked the solution $x=-3$ and evaluated $7x+8=-13$ for $x=-3$:

If we evaluate $7x+8$ for a different value of $x$, the left side will not be $-13$.

The equation $7x+8=-13$ is true when we replace the variable, $x$, with the value $-3$, but not true when we replace $x$ with any other value. Whether or not the equation $7x+8=-13$ is true depends on the value of the variable. Equations like this are called conditional equations. All the equations we have solved so far are conditional equations.

Now let’s consider the equation $2y+6=2(y+3)$. Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for $y$. Distribute: $2y+6=2y+6$. Subtract $2y$ to get the $y$’s to one side: $6=6$.

But $6=6$ is true. This means that the equation $2y+6=2(y+3)$ is true for any value of $y$. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.

What happens when we solve the equation $5z=5z-1$? Subtract $5z$ to get the constant alone on the right: $5z-5z=5z-5z-1$. Simplify — the $z$’s are gone: $0 \neq -1$.

But $0 \neq -1$. Solving the equation $5z=5z-1$ led to the false statement $0=-1$. The equation $5z=5z-1$ will not be true for any value of $z$. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Example. Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$6(2n-1)+3 = 2n-8+5(2n+1)$

Distribute: $12n-6+3=2n-8+10n+5$. Combine like terms: $12n-3=12n-3$. Subtract $12n$ to get the $n$’s to one side: $-3=-3$. This is a true statement. The equation is an identity. The solution is all real numbers.

Example. Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$10+4(p-5)=0$

Distribute: $10+4p-20=0$. Combine like terms: $4p-10=0$. Add $10$ to both sides: $4p=10$. Divide: $p=\tfrac{5}{2}$. The equation is true when $p=\tfrac{5}{2}$. This is a conditional equation. The solution is $p=\tfrac{5}{2}$.

Example. Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$5m+3(9+3m) = 2(7m-11)$

Distribute: $5m+27+9m=14m-22$. Combine like terms: $14m+27=14m-22$. Subtract $14m$ from both sides: $27 \neq -22$. But $27 \neq -22$. The equation is a contradiction. It has no solution.

Key terms

conditional equation — an equation that is true for one or more values of the variable and false for all other values. identity — an equation that is true for any value of the variable; its solution is all real numbers. contradiction — an equation that is false for all values of the variable; it has no solution.

This section is adapted from Elementary Algebra 2e, Section 2.4: Use a General Strategy to Solve Linear Equations by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the worked-example step tables as markdown tables; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback — including flipping the identity/contradiction classification Try Its into requests for the specific numerical statement each equation reduces to, since a word answer like “identity” can’t be graded by the math checker.