6x-2y+1=0 and 3x-5y+7=0

How do I substitute this?

thank you

shouldn't the 2 in 2(5y... be a 6?

no, 3x = 5y-7

So, 6x = 2(5y-7)

To solve a system of equations by substitution, follow these steps:

1. Choose one equation and solve it for one variable in terms of the other variable. Let's choose the first equation.

6x - 2y + 1 = 0

Simplify the equation by isolating one variable. For example, let's solve for x:

6x = 2y - 1

Divide both sides of the equation by 6:

x = (2y - 1) / 6

Now we have x in terms of y.

2. Substitute the expression found for x (or y) into the other equation. Let's substitute the expression we found for x in the second equation:

3x - 5y + 7 = 0

Replace x with (2y - 1) / 6:

3((2y - 1) / 6) - 5y + 7 = 0

Simplify the equation by distributing and combining like terms.

3. Solve the resulting equation for the remaining variable. In this case, we have an equation in terms of y. Simplify the equation and solve for y.

Multiply through by 6 to eliminate the fraction:

(2y - 1) - 30y + 42 = 0

Combine like terms:

-28y + 41 = 0

Subtract 41 from both sides:

-28y = -41

Divide both sides by -28:

y = (-41) / (-28)

Reduce the fraction if possible.

4. Substitute the value found for y back into either of the original equations to solve for the remaining variable. Let's use the first equation:

6x - 2((-41) / (-28)) + 1 = 0

Simplify and solve for x.

Substitute the values found for x and y into the equations and verify that they satisfy both equations.

That's how you can substitute equations to solve a system of equations using the substitution method.

from the 2nd equation, 3x = 5y-7

so the 1st equation becomes

2(5y-7)-2y+1 = 0

Now solve that for y, then figure x.