Math  /  Algebra

QuestionExercises Use substitution to solve each system of equations. 42y=13y=x+4\begin{array}{c} 4-2 y=-1 \\ 3 y=x+4 \end{array}
5. x4y=1x-4 y=1 2x8y=22 x-8 y=2
6. x+2y=0x+2 y=0 3x+4y=43 x+4 y=4

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx and yy that make both equations true at the same time! Watch out! Make sure you substitute correctly and keep track of your positive and negative signs!

STEP 2

1. Solve for x
2. Substitute
3. Solve for y
4. Final Answer

STEP 3

We're given 3y=x+43y = x + 4.
Let's **isolate** xx by adding 4-4 to both sides of the equation.
This is important because it gets xx all alone on one side of the equals sign! 3y4=x+443y - 4 = x + 4 - 4 3y4=x3y - 4 = xSo, x=3y4x = 3y - 4.

STEP 4

Now, let's **substitute** this expression for xx into the first equation, 42y=14 - 2y = -1.
This is where the magic happens!
We're replacing xx with something that equals xx, which is totally allowed! 42y=3y414 - 2y = 3y - 4 - 1

STEP 5

Let's **simplify** the right side: 42y=3y54 - 2y = 3y - 5

STEP 6

Now, let's get all the yy terms on one side.
We can do this by adding 2y2y to both sides: 42y+2y=3y5+2y4 - 2y + 2y = 3y - 5 + 2y 4=5y54 = 5y - 5

STEP 7

Next, let's **isolate** the constant term by adding 55 to both sides: 4+5=5y5+54 + 5 = 5y - 5 + 5 9=5y9 = 5y

STEP 8

Finally, let's **divide** both sides by 55 to get yy all by itself.
Remember, we want to make the coefficient of yy equal to **one**: 95=5y5\frac{9}{5} = \frac{5y}{5} 95=y\frac{9}{5} = ySo, y=95y = \frac{9}{5}.
Awesome!

STEP 9

Now that we know y=95y = \frac{9}{5}, we can **substitute** it back into the equation we found for xx: x=3y4x = 3y - 4. x=3954x = 3 \cdot \frac{9}{5} - 4 x=275205x = \frac{27}{5} - \frac{20}{5}x=75x = \frac{7}{5}

STEP 10

So, our **final answer** is x=75x = \frac{7}{5} and y=95y = \frac{9}{5}.
We did it!

STEP 11

x=75x = \frac{7}{5} and y=95y = \frac{9}{5}

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