Question

math 10c | unit 6 systems of linear equations
lesson 2 | solving systems algebraically
rf9 | solve problems that involve systems of linear equations in two variables, graphically & algebraically 9%
warm up | solve the following system graphically, verify your solution.
4x + y = 7
3x + 2y = -1
systems of linear equations |

• a system of linear equations is a group of 2 (or more) linear equations to be solved together

a solution to a system of equations is a point (x, y) that satisfies both equations

• this means it is a point common to both lines
• the intersection of the two lines

verifying a solution means you substitute the values from your point back into both equations & check to see if it makes the equation true (ls=rs)
solving systems algebraically | substitution
solving by substitution

1. solve one equation for x or y
2. substitute the expression found in step 1 into the other equation
3. solve for the remaining variable
4. use the answer from step 3 & sub into an equation to solve the other variable
5. state your solution as an ordered pair
6. verify your solution* (only necessary if question explicitly asks, however this is always good practice. you can verify algebraically or graphically)

Explanation:

Step1: Rewrite equations in slope - intercept form

For the equation \(4x + y=7\), solve for \(y\):
\(y=-4x + 7\). The \(y\) - intercept is \(7\) (when \(x = 0\), \(y = 7\)) and the \(x\) - intercept is found by setting \(y = 0\): \(0=-4x + 7\), \(4x=7\), \(x=\frac{7}{4}=1.75\) (or \((\frac{7}{4},0)\)).

For the equation \(3x + 2y=-1\), solve for \(y\):
\(2y=-3x - 1\), \(y =-\frac{3}{2}x-\frac{1}{2}\). The \(y\) - intercept is \(-\frac{1}{2}\) (when \(x = 0\), \(y=-\frac{1}{2}\)) and the \(x\) - intercept is found by setting \(y = 0\): \(0=-\frac{3}{2}x-\frac{1}{2}\), \(\frac{3}{2}x=-\frac{1}{2}\), \(x=-\frac{1}{3}\) (or \((-\frac{1}{3},0)\)).

Step2: Graph the two lines

• For \(y=-4x + 7\), plot the \(y\) - intercept \((0,7)\) and the \(x\) - intercept \((\frac{7}{4},0)\) and draw the line.
• For \(y =-\frac{3}{2}x-\frac{1}{2}\), plot the \(y\) - intercept \((0,-\frac{1}{2})\) and the \(x\) - intercept \((-\frac{1}{3},0)\) and draw the line.

The point of intersection of the two lines is the solution of the system. By graphing (or we can also solve algebraically to verify), we can find the intersection point. Let's solve algebraically using substitution. From \(4x + y=7\), we have \(y = 7 - 4x\). Substitute \(y = 7 - 4x\) into \(3x+2y=-1\):

\(3x + 2(7 - 4x)=-1\)

Step3: Solve the substituted equation

Expand the left - hand side: \(3x+14 - 8x=-1\)

Combine like terms: \(-5x+14=-1\)

Subtract \(14\) from both sides: \(-5x=-1 - 14=-15\)

Divide both sides by \(-5\): \(x = 3\)

Step4: Find the value of \(y\)

Substitute \(x = 3\) into \(y = 7 - 4x\): \(y=7-4\times3=7 - 12=-5\)

Step5: Verify the solution

• For the equation \(4x + y=7\):

Left - hand side (LHS): \(4\times3+(-5)=12 - 5 = 7\), Right - hand side (RHS): \(7\). So \(LHS = RHS\).

• For the equation \(3x + 2y=-1\):

Left - hand side (LHS): \(3\times3+2\times(-5)=9-10=-1\), Right - hand side (RHS): \(-1\). So \(LHS = RHS\).

Answer:

The solution of the system \(

$$\begin{cases}4x + y=7\\3x + 2y=-1\end{cases}$$

\) is \((3,-5)\)