Question

math 10c | unit 6 systems of linear equations
lesson 1 | solving systems graphically
9 | solve problems that involve systems of linear equations in two variables, graphically & algebraically 9%
warm up | solve each of the following equations. verify that your solution satisfies the equation.
a. ( m = 26.8 - 6.8 + 3m )
b. ( 4(y + 2) - 5(y + 1) = y - 1 )

Explanation:

Step1: Solve for \( m \) in \( m = 26.8 - 6.8 + 3m \)

Simplify the right - hand side: \( 26.8-6.8 = 20 \), so the equation becomes \( m=20 + 3m \).
Subtract \( 3m \) from both sides: \( m-3m=20+3m - 3m \), which gives \( - 2m=20 \).
Divide both sides by - 2: \( m=\frac{20}{-2}=- 10 \).

Step2: Verify the solution

Substitute \( m = - 10 \) into the original equation \( m = 26.8-6.8 + 3m \).
Left - hand side (LHS): \( m=-10 \).
Right - hand side (RHS): \( 26.8-6.8+3\times(-10)=20 - 30=-10 \).
Since \( LHS = RHS=-10 \), the solution is correct.

Step1: Solve for \( y \) in \( 4(y + 2)-5(y + 1)=y - 1 \)

Expand the brackets: \( 4y+8-5y - 5=y - 1 \).
Simplify the left - hand side: \( (4y-5y)+(8 - 5)=-y + 3 \). So the equation is \( -y + 3=y - 1 \).
Add \( y \) to both sides: \( -y + y+3=y + y-1 \), which gives \( 3 = 2y-1 \).
Add 1 to both sides: \( 3 + 1=2y-1 + 1 \), so \( 4 = 2y \).
Divide both sides by 2: \( y = 2 \).

Step2: Verify the solution

Substitute \( y = 2 \) into the original equation \( 4(y + 2)-5(y + 1)=y - 1 \).
Left - hand side (LHS): \( 4\times(2 + 2)-5\times(2 + 1)=4\times4-5\times3=16 - 15 = 1 \).
Right - hand side (RHS): \( 2-1 = 1 \).
Since \( LHS = RHS = 1 \), the solution is correct.

Answer:

For part (a), \( m=-10 \); for part (b), \( y = 2 \)