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Study over this website

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There is no doubt that solving a two-step equation is extremely easy. As the name implies, two-step equations can be solved in just two steps. If this is your first encounter with two-step equations, don't worry because we will go over enough examples to make you familiar with the process.

When solving an equation in general, we always keep in mind the notion that whatever we do to one side of the equation should also be done to the other side to ensure that the equation remains balanced.

We know that we have completely solved a two-step equation if the variable, usually represented by a letter in an alphabet, is isolated on one side of the equation (either left or right) and the number is located on the opposite side.

The USUAL way of solving a two-step equation:

Note: This is the "usual" method because most of the two-step equations are solved this way. Notice that Step 2 can alternatively be replaced by Step 3 which are the same essentially.

1) First, add or subtract both sides of the linear equation by the same number.

2) Secondly, multiply or divide both sides of the linear equation by the same number.

3)* Instead of step #2, always multiply both sides of the equation by the reciprocal of the coefficient of the variable.

Examples of How to Solve Two-Step Equations

Example 1: Solve the two-step equation below.

As the name of this linear equation suggests, it requires two steps in order to solve for the unknown variable. Generally, the first step involves getting rid of the number "farthest" from the term with a variable being solved. Then, we eliminate the number "closest" to the variable. The number is either multiplying or dividing the variable. It is also called the coefficient of the term.

The variable here is [latex]x[/latex]. Our goal is to solve [latex]x[/latex] by isolating it on one side of the equation. Keeping the variable on the left or on the right doesn't make any difference. It is up to you! In this problem, let's keep it on the left side since it's already there.

On the side (left side of the linear equation) where the variable is located, notice that [latex]2[/latex] is "closest" to variable [latex]x[/latex], and [latex]5[/latex] is "farthest" one.

This simple observation allows us to decide which number to eliminate first. It is obviously [latex]+5[/latex] because it is farther between the two. The opposite of [latex]+5[/latex] is [latex]-5[/latex], that means we will subtract both sides of the equation by [latex]5[/latex].

After eliminating [latex]5[/latex] on the left side of the equation by subtracting both sides by [latex]5[/latex], it's time to get rid of the number closest or directly attached to [latex]x[/latex] which is [latex]2[/latex] in [latex]2x[/latex]. Since [latex]2[/latex] is multiplying the variable [latex]x[/latex], its opposite operation is to divide by [latex]2[/latex].

After dividing both sides by [latex]2[/latex], we obtain the final answer or solution to the given two-step linear equation.

Just a quick reminder, it is considered solved because the coefficient of the variable is just positive one, [latex]+1[/latex].

Example 2: Solve the two-step equation below.

Our goal is to keep the variable [latex]x[/latex] on one side of the equation. It doesn't matter which side, however it is a "standard" practice to keep the variable being solved to the left side. Some algebra teachers may require you to keep the variable to the left and there's nothing with that. Personally, I don't mind where you keep the variable, either left or right, as long as the isolated variable on one side of the equation has a coefficient of [latex]+1[/latex].

The first step involves removing the number "farthest" from variable [latex]x[/latex]. Notice that [latex]-3[/latex] is "closest" to [latex]x[/latex], while [latex]-8[/latex] is "farther away". So then, we can eliminate [latex]-8[/latex] by adding to its opposite which is [latex]+8[/latex].

The second step involves getting rid of the number closest to the variable [latex]x[/latex] which is [latex]-3[/latex]. Since [latex]-3[/latex] is multiplying the variable x, its opposite operation is to divide by [latex]-3[/latex]. After dividing both sides by [latex]-3[/latex], we have solved the linear equation.

Quick reminder, [latex]-3[/latex] divided by [latex]-3[/latex] is equal to [latex]+1[/latex].

Example 3: Solve the two-step equation below.

Here's a situation where we can isolate the variable [latex]x[/latex] to the right side of the equation since it is already there.

Looking at the right side of the equation where the variable is located, the number [latex]3[/latex] is closest to [latex]x[/latex] because [latex]3[/latex] is dividing the variable [latex]x[/latex]. On the other hand, the number [latex]26[/latex] is "farther away". This implies that we will have to deal with [latex]+26[/latex] by subtracting both sides of the equation by [latex]26[/latex]. The reason we subtract is that the additive inverse of [latex]+26[/latex] is [latex]-26[/latex].

The second step is to get rid of the denominator [latex]3[/latex]. Since [latex]3[/latex] is dividing the [latex]x[/latex], its opposite operation is to multiply by [latex]3[/latex].

After multiplying both sides by [latex]3[/latex], we have arrived at the final answer. You may rewrite your final answer as [latex]x = -9[/latex].

Example 4: Solve the two-equation below.

This may appear to be a multi-step equation, but it is not. It can be solved in two steps. Don't be bothered by the fractions because they are very easy to deal with. In this case, you will apply the rule of fraction addition. The rule states that if you are adding two fractions that have the same denominator, just add the numerators then copy the common denominator.

Back to solving the two-step equation above, to remove the fraction on the left side which is [latex]\Large{ - {3 \over {10}}}[/latex], we will add [latex]\Large{{3 \over {10}}}[/latex] to both sides of the equation.

The reason that we are adding instead of subtracting is because the additive inverse of [latex]\Large{ - {3 \over {10}}}[/latex] is [latex]\Large{+ {3 \over {10}}}[/latex] .

After adding [latex]\Large{{3 \over {10}}}[/latex] to both sides, only [latex]{\Large{{2 \over 5}}}x[/latex] remains on the left side.

For the right side of the equation, we have [latex]\Large{{9 \over {10}} + {3 \over {10}} = {{12} \over {10}}}[/latex].

Everything I stated above is just the first step. Now, moving on to the second step. Look at the coefficient of the variable [latex]x[/latex]. It is [latex]\Large{{2 \over 5}}[/latex] that means its reciprocal is [latex]\Large{{5 \over 2}}[/latex].

To finally solve the given equation, we will multiply both sides of the equation by the reciprocal of the coefficient of the variable in question. Here's the complete step-by-step solution:

You may also be interested in these related math lessons or tutorials: Two-Step Equations Practice Problems with Answers Multi-Step Equations Practice Problems with Answers

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Solving Two-Step Equations There is no doubt that solving a two-step equation is extremely easy. As the name implies, two-step equations can be solved in just two steps. If this is your first encounter with two-step equations, don't worry because we will go over enough examples to make you familiar with the process.

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Watch carefully!

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Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them. Now, we are going to figure out what X is. But before we even do that, I want you to think about a mathematical equation that can represent what is going on; that equates what we have on the left hand, with what we have on the right side of the scale. I will give you a few seconds to think about it... So let's think about what we have on the left side: we have 3 masses with mass X, so you can say we have 3x and then we have 2 masses of 1 kilogram, so in total we have 2 kg. So + 2. So one way to think about the total mass on the left-hand side is 3x + 2. Three masses with mass X, plus two kilograms. That is what we have on the left-hand side. Now, let us think about what we have on the right-hand side. We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: you can take these 1kg blocks off of the left-hand side, but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, our scale is still balanced. And our equation: 3x is now equal to 12. Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. If we want to keep a third of what we had here originally, -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced. So you see that the mass of this thing must be the same as the mass of these 4 left-over blocks. x must be equal to 4 kilograms.

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Intro to two-step equations | Algebra (video) | Khan Academy

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Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them.

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assignment 3

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2-Step Equations Worksheet and Solutions

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Math Worksheet: How to solve equations requiring two operations worksheets and solutions

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Assignment 3