- Welcome to this lesson in which we are going to talk about solving linear equations, and we're going to cover the core equation solving skills you need for SAT algebra. The goal of solving a linear equation is simple.

Get the variable all by itself on one side of the equation. And there's one big rule that makes everything work. Whatever you do to one side of the equation, you must also do to the other side. That keeps the equation balanced. So solving equations is basically undoing operations step by step.

Let's try one. 3x plus 4 equals 19. Our goal is to isolate x. Right now, four is being added to 3x, so let's undo that first. We undo that by subtracting 4 from both sides, which is going to leave us with 3x equals 15.

Now, x is being multiplied by 3, and we undo that by dividing both sides by 3. 3x divided by 3 equals 15 divided by 3, and that's going to simplify 2x equals 5. We're done. Now, let's look at a very common SAT variation. Variables on both sides. Check out this equation.

5x plus 7 equals 2x minus 5. Looks a little scarier, but the strategy is the same. First, get all the x terms onto one side. The smaller variable term is 2x. So let's subtract 2x from both sides. That's going to leave us with 3x plus 7 equals -5.

Now, this is just a regular two-step equation. We're going to subtract 7 from both sides. We'll have 3x minus 12 after we do so. And then we are going to divide both sides by 3, and that's going to leave us with x equals a -4. One common mistake here is subtracting the variable term from only one side.

Remember, every operation must happen on both sides of the equation. All right, now let's talk about everyone's favorite subject, fractions. Suppose we have 2/3x equals 6. A lot of students instinctively cross-multiply here, but there's a faster move.

Multiply both sides by the reciprocal of 2/3rds. The reciprocal of 2/3rds is found by just flipping the fraction over. So the reciprocal of 2/3rds is 3/2. Now let's apply that to our problem. 3/2 multiplied by 2/3x equals 6 times 3/2.

On the left, the fractions cancel. On the right, we have 6 times 3/2 equals 9. So that's going to leave us with x equals 9. Super fast and super clean. Now, let's look at a different fraction situation. Fractions being added or subtracted.

Let's solve the following. 7x/6 plus 2/3rds equals 13/2. When fractions are added or subtracted, the fastest strategy is usually clearing out the fractions. Because if you have a choice between fractions and integers, integers is probably the better way to go, especially in terms of efficiency.

To clear them out, we're going to multiply every term by the least common multiple, or LCM, of the denominators. The denominators here are 3, 6, and 2. The lowest common multiple is 6. So we're going to multiply every term by 6. That's 6 times 7x/6 plus 6 times 2/3rds equals 6 times 13/2.

Now we're going to work through them. The 6s cancel in the first term, leaving us with just 7x. Next, the second term, we've got 6 times 2/3rds. That's going to give us 4. And on the other side of the equal sign, we've got 6 times 13/2.

That's going to give us 39. The fractions are gone. Our new equation is 7x plus 4 equals 39. Now we just solve like normal. We're going to subtract 4 from both sides and have 7x equals 35. And then we're going to divide both sides by 7 and we can see that x equals 5.

This is one of the highest leverage SAT algebra moves, clear fractions using the LCM. Now let's talk about a sneaky SAT pattern, solving for an expression instead of solving directly for x. Here's the problem. 2x plus 5 equals 19. What is the value of 2x plus 7?

Notice the question is not asking for x. And sometimes you can answer much faster if you solve directly for the expression you want. Look at what we already have. We've got 2x plus 5 equals 19. We want 2x plus 7. We have to think, "Okay, "how do I turn 2x plus 5 into 2x plus 7?" We do that by adding 2 to both sides.

Add 2 to both sides of the equal sign and simplify it, and we are already at 2x plus 7 equals 21. So the answer is 21. We never solve for x. We didn't need to. This is a time saver, so always reread the question carefully and make sure you're solving for the quantity actually being asked.

Two quick footnotes before we wrap up. Sometimes the variables cancel completely and you end up with a false statement, like if you have 5 equals 7. When that happens, the equation has no solution. Other times, the variables cancel and you get a true statement like 4 equals 4.

That means every value would work, so the equation has infinitely many solutions. We'll go deeper into those cases later when we talk about systems of equations. But for now, we're going to recap. When solving linear equations, isolate the variable.

Do the same thing to both sides of the equal sign. Move variables onto one side when they appear on both sides. Use reciprocals to clear fractions multiplied by variables. Use the LCM, or lowest common multiple, to clear added or subtracted fractions, and always double check what the question is actually asking.

That's it for solving linear equations. Up next will be linear equations in word problems. Thanks for watching.